I can't understand logical implication
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I just started studying logic (high school)
anyway...for the truth table of logical implication
If sentence $A$ is true and $B$ is true then $Aimplies B$ is true.
does that mean if $A$ and $B$ are both true
then there is a way to prove $B$ is true from $A$,
always?
the same for if $A$ is false can you get anything either True or false proved from this $A$?
logic intuition propositional-calculus
edited Sep 27 '14 at 10:06
Hanul Jeon
17.5k42780
asked Sep 27 '14 at 10:03
linalina
96114
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|
show 13 more comments
$begingroup$
I just started studying logic (high school)
anyway...for the truth table of logical implication
If sentence $A$ is true and $B$ is true then $Aimplies B$ is true.
does that mean if $A$ and $B$ are both true
then there is a way to prove $B$ is true from $A$,
always?
the same for if $A$ is false can you get anything either True or false proved from this $A$?
logic intuition propositional-calculus
edited Sep 27 '14 at 10:06
Hanul Jeon
17.5k42780
asked Sep 27 '14 at 10:03
linalina
96114
$endgroup$
10
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@gebruiker Yes, you can.
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– Git Gud
Sep 27 '14 at 10:10
3
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@gebruiker - Why would you think you can't ?
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– Belgi
Sep 27 '14 at 10:20
3
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@Belgi Because I own a dog is true, and I am a man is true. But owning a dog does not imply being a man. I know women who own dogs too. Or am I overlooking something here?
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– gebruiker
Sep 27 '14 at 10:32
10
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@gebruiker "(gebruiker owns a dog) implies (gebruiker is a man)" is true. "(X owns a dog) implies (X is a man)" is not true for all X
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– Ben Aaronson
Sep 27 '14 at 20:33
6
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$Aimplies B ;;;;$ means $;;; operatorname{truthvalue}(A) leq operatorname{truthvalue}(B) ;;;$. $;;;;;;;$
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– user57159
Sep 28 '14 at 2:01
|
show 13 more comments
$begingroup$
I just started studying logic (high school)
anyway...for the truth table of logical implication
If sentence $A$ is true and $B$ is true then $Aimplies B$ is true.
does that mean if $A$ and $B$ are both true
then there is a way to prove $B$ is true from $A$,
always?
the same for if $A$ is false can you get anything either True or false proved from this $A$?
logic intuition propositional-calculus
edited Sep 27 '14 at 10:06
Hanul Jeon
17.5k42780
asked Sep 27 '14 at 10:03
linalina
96114
$endgroup$
I just started studying logic (high school)
anyway...for the truth table of logical implication
If sentence $A$ is true and $B$ is true then $Aimplies B$ is true.
does that mean if $A$ and $B$ are both true
then there is a way to prove $B$ is true from $A$,
always?
the same for if $A$ is false can you get anything either True or false proved from this $A$?
logic intuition propositional-calculus
logic intuition propositional-calculus
edited Sep 27 '14 at 10:06
Hanul Jeon
17.5k42780
asked Sep 27 '14 at 10:03
linalina
96114
edited Sep 27 '14 at 10:06
Hanul Jeon
17.5k42780
asked Sep 27 '14 at 10:03
linalina
96114
edited Sep 27 '14 at 10:06
Hanul Jeon
17.5k42780
edited Sep 27 '14 at 10:06
Hanul Jeon
17.5k42780
edited Sep 27 '14 at 10:06
Hanul Jeon
17.5k42780
17.5k42780
asked Sep 27 '14 at 10:03
linalina
96114
asked Sep 27 '14 at 10:03
linalina
96114
asked Sep 27 '14 at 10:03
linalina
96114
96114
10
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@gebruiker Yes, you can.
$endgroup$
– Git Gud
Sep 27 '14 at 10:10
3
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@gebruiker - Why would you think you can't ?
$endgroup$
– Belgi
Sep 27 '14 at 10:20
3
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@Belgi Because I own a dog is true, and I am a man is true. But owning a dog does not imply being a man. I know women who own dogs too. Or am I overlooking something here?
$endgroup$
– gebruiker
Sep 27 '14 at 10:32
10
$begingroup$
@gebruiker "(gebruiker owns a dog) implies (gebruiker is a man)" is true. "(X owns a dog) implies (X is a man)" is not true for all X
$endgroup$
– Ben Aaronson
Sep 27 '14 at 20:33
6
$begingroup$
$Aimplies B ;;;;$ means $;;; operatorname{truthvalue}(A) leq operatorname{truthvalue}(B) ;;;$. $;;;;;;;$
$endgroup$
– user57159
Sep 28 '14 at 2:01
|
show 13 more comments
10
$begingroup$
@gebruiker Yes, you can.
$endgroup$
– Git Gud
Sep 27 '14 at 10:10
3
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@gebruiker - Why would you think you can't ?
$endgroup$
– Belgi
Sep 27 '14 at 10:20
3
$begingroup$
@Belgi Because I own a dog is true, and I am a man is true. But owning a dog does not imply being a man. I know women who own dogs too. Or am I overlooking something here?
$endgroup$
– gebruiker
Sep 27 '14 at 10:32
10
$begingroup$
@gebruiker "(gebruiker owns a dog) implies (gebruiker is a man)" is true. "(X owns a dog) implies (X is a man)" is not true for all X
$endgroup$
– Ben Aaronson
Sep 27 '14 at 20:33
6
$begingroup$
$Aimplies B ;;;;$ means $;;; operatorname{truthvalue}(A) leq operatorname{truthvalue}(B) ;;;$. $;;;;;;;$
$endgroup$
– user57159
Sep 28 '14 at 2:01
10
10
$begingroup$
@gebruiker Yes, you can.
$endgroup$
– Git Gud
Sep 27 '14 at 10:10
$begingroup$
@gebruiker Yes, you can.
$endgroup$
– Git Gud
Sep 27 '14 at 10:10
3
3
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@gebruiker - Why would you think you can't ?
$endgroup$
– Belgi
Sep 27 '14 at 10:20
$begingroup$
@gebruiker - Why would you think you can't ?
$endgroup$
– Belgi
Sep 27 '14 at 10:20
3
3
$begingroup$
@Belgi Because I own a dog is true, and I am a man is true. But owning a dog does not imply being a man. I know women who own dogs too. Or am I overlooking something here?
$endgroup$
– gebruiker
Sep 27 '14 at 10:32
$begingroup$
@Belgi Because I own a dog is true, and I am a man is true. But owning a dog does not imply being a man. I know women who own dogs too. Or am I overlooking something here?
$endgroup$
– gebruiker
Sep 27 '14 at 10:32
10
10
$begingroup$
@gebruiker "(gebruiker owns a dog) implies (gebruiker is a man)" is true. "(X owns a dog) implies (X is a man)" is not true for all X
$endgroup$
– Ben Aaronson
Sep 27 '14 at 20:33
$begingroup$
@gebruiker "(gebruiker owns a dog) implies (gebruiker is a man)" is true. "(X owns a dog) implies (X is a man)" is not true for all X
$endgroup$
– Ben Aaronson
Sep 27 '14 at 20:33
6
6
$begingroup$
$Aimplies B ;;;;$ means $;;; operatorname{truthvalue}(A) leq operatorname{truthvalue}(B) ;;;$. $;;;;;;;$
$endgroup$
– user57159
Sep 28 '14 at 2:01
$begingroup$
$Aimplies B ;;;;$ means $;;; operatorname{truthvalue}(A) leq operatorname{truthvalue}(B) ;;;$. $;;;;;;;$
$endgroup$
– user57159
Sep 28 '14 at 2:01
|
show 13 more comments
11 Answers
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As a logical proposition, the material conditional $A implies B$ is a very weak one: as you've noticed, it's very easy to satisfy it just by accident. In fact, this happens whenever $A$ is false, or whenever $B$ is true. Thus, merely observing that $A implies B$, for some specific $A$ and $B$, says very little.
Instead, the usefulness of implication lies in the fact that, precisely because of its weakness, it is often possible to assert $A implies B$ as a universal statement (either an axiom or a provable theorem) that holds for any valuation of any free variables mentioned in the propositions $A$ and $B$.
For example, consider the statement: $$x > 2 ;land; x text{ is prime} implies x text{ is odd}.$$ Merely observing that this statement holds for some $x$ says very little — there are plenty of numbers for which it is trivially true, either because they are odd, or because they are not primes greater than 2. What makes this statement useful is that we can prove that it holds for all $x$ — there isn't a single number which would be greater than 2 and prime, but not odd.
answered Sep 27 '14 at 14:23
Ilmari KaronenIlmari Karonen
19.6k25183
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This is the answer that gets to the heart of the matter. +1.
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– 6005
Sep 27 '14 at 16:18
add a comment |
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One way to understand implication is to remember that $ARightarrow B$ is equivalent to $neg A lor B$. If you understand negation ($neg$) and disjunction ($lor$), then you understand implication.
answered Sep 27 '14 at 20:30
Theodore NorvellTheodore Norvell
39119
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I thought this was a strange equivalence, until I noticed myself saying things like: "I didn't know that or I would have done this"
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– Justin
Sep 28 '14 at 4:47
1
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I find $neg[Aland neg B]$ easier to work with.
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– Dan Christensen
Sep 29 '14 at 15:39
add a comment |
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Look at $A$ and $B$ as something that is either false or true. For
example let $A$ be the event that tomorrow is Tuesday and let $B$
be the event that the day after tomorrow is Wednesday.
Look at
$$
Aimplies B
$$
as a promise - if $A$ is true then so is $B$.
In our example, if $A$ is true then indeed so is $B$ and so the
implication $Aimplies B$ is true.
However, now consider $C$ as the statement that tomorrow is Friday,
and I state
$$
Cimplies B
$$
that is - I promise you that if $C$ will happen so will $B$.
Tomorrow is not Friday (at the time of writing), and so $C$ is false,
regardless of if $B$ is false or true - my promise was kept.
Now regarding the terminology of proofing $B$ is both $A,B$ are
true. Note that statements like
$$
text{My cat walks on four}implies1+1=2
$$
is true, since both are true, but what would it mean to prove $B$ from $A$
?
answered Sep 27 '14 at 10:18
BelgiBelgi
14.5k954113
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I would say that $A$ being true and $B$ being true does not mean you can always prove (deduce) $B$ from $A$.
Here's an example. A: Alice lives in Atlanta. B: Bob lives in Boston. Even if these are both true, there is no (apparent) relationship. So you can't logically deduce $B$ from $A$ even though $ARightarrow B$ is true in this case.
I guess this means that if you can logically deduce statement $Q$ from statement $P$, then $PRightarrow Q$ is true; but knowing $PRightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption of $P$.
Logical implication is a defined logical connective, so as long as $P$ and $Q$ have truth values (true or false), so does $PRightarrow Q$.
answered Sep 27 '14 at 10:17
paw88789paw88789
29.1k12349
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You can logically deduce B from A in that example. Let's assume A is true. B is true. QED. Yes, this counts, unless you really meant A implies B and also not not-A implies not-B. But if you meant that, you should have said so.
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– djechlin
Sep 28 '14 at 2:00
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Your 3rd paragraph is incorrect. Suppose $PRightarrow Q$ is given, we construct a proof of $Q$ to $P$: Assume $P$, assume $neg Q$, then by the truth table for $PRightarrow Q$ (or by modes ponens) $Q$. Hence $Q$ and $neg Q$, a contradiction. Hence our assumption $neg Q$ was false.
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– James
Sep 29 '14 at 16:41
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My point is that sometimes two statements, $P$ and $Q$ are unrelated. Now $PRightarrow Q$ can perfectly well be true (provided $P$ is false or $Q$ is true), and yet knowing that $P$ is true does not help us determine (deduce) that $Q$ is true, because the statements are unrelated.
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– paw88789
Sep 29 '14 at 17:43
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@paw88789 But when you know $P$, then to quote you "knowing $PRightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption that $P$." This is wrong, I just gave a deduction.
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– James
Sep 29 '14 at 18:42
add a comment |
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Maybe it's more clear if we separate the logical operator meaning of implication from its logical statement meaning.
When we use it as a logical operator, we conceive it simply as an entity, that given two logical values (thus true or false), produces a third logical value, using a common defined rule (its truth table).
So it makes perfect sense to say $A implies B = true$ if $A = B = true$, and we don't concern about what actually proposition $A$ and $B$ means, we care exclusively about their logical values.
It's different when we use implication as a logical statement.
In this case we really say something about the meaning of the propositions involved in our statement. So, while proposition $A = My;cat;is;black$ is true, and proposition $B = I;am;hungry$ is true as well, $A implies B$ is not a valid statement.
Such meaning is linked to set theory and formal logic. Using implication in this context means that you can infer $B$ from $A$, in a way called modus ponens.
answered Sep 27 '14 at 11:11
mattecapumattecapu
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Implication can be understood as thinking in NECESSARY and SUFFICIENT conditions. A good example is to think about someone who has born in Dallas, Texas.
Proposition P could stand for "Someone who was born in Dallas"
Proposition Q could stand for "Someone who is texan"
Truth table for implication is:
So, in this example is easy to check line by line the validity of truth table above:
Who was born in Dallas is therefore texan; So the first implication is TRUE.
Who was born in Dallas MUST be texan. So, saying that someone who was born in Dallas, in the state of Texas, and is not texan is FALSE;
Saying who wasn't born in Dallas but is texan anyway (take someone who has born in Houston, for example) is a TRUE statement;
At last, who wasn't born in Dallas AND also is NOT texan is also TRUE.
So, thinking in terms of NECESSARY (to be texan is a necessary condition for who was born in Dallas) and SUFFICIENT (to born in Dallas is a sufficient condition to be texan) conditions ease the process to translate implication to natural language.
Other examples of propositions containing necessary and sufficient conditions are:
- P = To be a priest / Q = To believe in God;
- P = To be a Police officer / Q = To carry a gun;
- P = The number is divisible by 4 / Q = The number is even.
answered Mar 30 '15 at 9:37
Daniel LavieriDaniel Lavieri
213
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The problem is understanding that there's a difference between the everyday mathematical meaning of "implies" - "I can prove it (or someone can)" - and the formal logical meaning expressed by the truth table. This may help: In classical logic, why is $(pRightarrow q)$ True if $p$ is False and $q$ is True?.
edited Apr 13 '17 at 12:21
Community♦
1
answered Sep 27 '14 at 22:22
Ethan BolkerEthan Bolker
42.1k548111
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If sentence $A$ is true and $B$ is true then $Aimplies B$ is true.
Yes. In mathematics and logic as widely applied, $Aimplies B$ is equivalent to $neg[Alandneg B]$.
So, if $A$ is true and $B$ is true, then $neg[Alandneg B]$ and hence $Aimplies B$ will also be true.
Does that mean if $A$ and $B$ are both true then there is a way to prove $B$ is true from $A$, always?
Yes. Here is an informal proof:
Suppose $A land B$.
Suppose $A$.
$B$ (from 1)
$Aimplies B$ (conclusion from 2 and 3)
$Aland B implies [Aimplies B]$ (conclusion from 1 and 4)
Alternatively, prove (5) using truth table.
The same for if $A$ is false can you get anything either True or false proved from this $A$?
Yes. Informal proof:
Suppose $neg A$
Suppose $Aland neg B$
$Aland neg A$ (contradiction from 1 and 2)
$neg[Aland neg B]$ (by contradiciton from 2 and 3)
$Aimplies B$ from (equivalently from 4)
$neg Aimplies [Aimplies B] $ (conclusion from 1 and 5)
Alternatively, prove (6) using a truth table.
That the above may seem counter-intuitive may stem from the fact that implication is often confused with temporal notions of causality. The statement, "If it is raining, then it is cloudy" on its own does not mean that rain always causes clouds, or that clouds always cause rain. If means only that, at a given instant in time, its is not the case that it is raining and not cloudy.
answered Sep 29 '14 at 15:55
Dan ChristensenDan Christensen
8,55321833
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The sentence $Aimplies B$ is not a sentence about proving anything. It is a compound sentence, but otherwise just like the sentences $A$ and $B$. For the truth table, the compound sentence got to have truth values like $A$ and $B$ has, and it has to depend on these sentences truth values. The values for this compound sentence is the same as for the compound statement $neg A vee B$.
answered Sep 29 '14 at 19:14
LehsLehs
6,98031662
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This is the way I understand it (I make no guarantee this is considered a correct interpretation).
the truth value of an implies statement is basically 'innocent until proven guilty', or 'my statement is true so long as the proposition values don't disprove it'.
So with an example, propositions P and Q, with the statement P implies Q. Lets say proposition P is 'is a man' and proposition Q is 'lives in japan', so our implication statement is basically saying 'all men live in japan'.
The truth table:
is basically saying, for any example where P and Q are true, we haven't disproved our statement that P -> Q, and for any example where P is false we haven't disproved our statement, only if we have an example of P being true and Q being false is our statement shown to be false.
So imagine there are only 4 humans in existence, we look at each of them and categorise them in our truth table. By doing this we can work out if our statement that P -> Q is true. We find 2 of the people are men, and that they do live in Japan, OK good, our statement still holds true, the 3rd person is a woman, we don't care if she lives in Japan or not because in either case our statement about men is still holding true.
A woman living in or outside japan has no bearing on the truth of our statement about men living in japan.
The 4th person then will decide if our statement is true or not, if the 4th person is a man living in japan, or a woman, then is it logically true to say that being a man implies that you live in japan.
Why is this useful? well the truth table itself doesn't help us prove the truth of our statement by itself, what it does is clearly set out what conditions are relevant to the truth of our statement - i.e. that we don't care about women, and that only a man not living in japan will falsify our statement.
Of course, you could make a statement like 'all green men live on mars'. This is a true statement because we cannot verify who lives on mars, and we don't have any examples of green men. But this is whats called a "vacuous truth", it has no bearing on reality for us because the propositions are intangible.
answered Nov 22 '17 at 9:13
Coder375Coder375
101
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This illustration of logical implication might help:
a) When you truly understand logical implication,
b) then you’ll be a happy person.
You may be a happy person for other reasons. But if you’re unhappy, then certainly you don’t truly understand logical implication. :-)
Truth table:
a b a⇒b
F F T
F T T
T F F
T T T
I hope this somewhat intuitive example helps.
answered Jan 6 at 5:52
Bert SierraBert Sierra
12
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11 Answers
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11 Answers
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As a logical proposition, the material conditional $A implies B$ is a very weak one: as you've noticed, it's very easy to satisfy it just by accident. In fact, this happens whenever $A$ is false, or whenever $B$ is true. Thus, merely observing that $A implies B$, for some specific $A$ and $B$, says very little.
Instead, the usefulness of implication lies in the fact that, precisely because of its weakness, it is often possible to assert $A implies B$ as a universal statement (either an axiom or a provable theorem) that holds for any valuation of any free variables mentioned in the propositions $A$ and $B$.
For example, consider the statement: $$x > 2 ;land; x text{ is prime} implies x text{ is odd}.$$ Merely observing that this statement holds for some $x$ says very little — there are plenty of numbers for which it is trivially true, either because they are odd, or because they are not primes greater than 2. What makes this statement useful is that we can prove that it holds for all $x$ — there isn't a single number which would be greater than 2 and prime, but not odd.
answered Sep 27 '14 at 14:23
Ilmari KaronenIlmari Karonen
19.6k25183
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$begingroup$
This is the answer that gets to the heart of the matter. +1.
$endgroup$
– 6005
Sep 27 '14 at 16:18
add a comment |
$begingroup$
As a logical proposition, the material conditional $A implies B$ is a very weak one: as you've noticed, it's very easy to satisfy it just by accident. In fact, this happens whenever $A$ is false, or whenever $B$ is true. Thus, merely observing that $A implies B$, for some specific $A$ and $B$, says very little.
Instead, the usefulness of implication lies in the fact that, precisely because of its weakness, it is often possible to assert $A implies B$ as a universal statement (either an axiom or a provable theorem) that holds for any valuation of any free variables mentioned in the propositions $A$ and $B$.
For example, consider the statement: $$x > 2 ;land; x text{ is prime} implies x text{ is odd}.$$ Merely observing that this statement holds for some $x$ says very little — there are plenty of numbers for which it is trivially true, either because they are odd, or because they are not primes greater than 2. What makes this statement useful is that we can prove that it holds for all $x$ — there isn't a single number which would be greater than 2 and prime, but not odd.
answered Sep 27 '14 at 14:23
Ilmari KaronenIlmari Karonen
19.6k25183
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$begingroup$
This is the answer that gets to the heart of the matter. +1.
$endgroup$
– 6005
Sep 27 '14 at 16:18
add a comment |
$begingroup$
As a logical proposition, the material conditional $A implies B$ is a very weak one: as you've noticed, it's very easy to satisfy it just by accident. In fact, this happens whenever $A$ is false, or whenever $B$ is true. Thus, merely observing that $A implies B$, for some specific $A$ and $B$, says very little.
Instead, the usefulness of implication lies in the fact that, precisely because of its weakness, it is often possible to assert $A implies B$ as a universal statement (either an axiom or a provable theorem) that holds for any valuation of any free variables mentioned in the propositions $A$ and $B$.
For example, consider the statement: $$x > 2 ;land; x text{ is prime} implies x text{ is odd}.$$ Merely observing that this statement holds for some $x$ says very little — there are plenty of numbers for which it is trivially true, either because they are odd, or because they are not primes greater than 2. What makes this statement useful is that we can prove that it holds for all $x$ — there isn't a single number which would be greater than 2 and prime, but not odd.
answered Sep 27 '14 at 14:23
Ilmari KaronenIlmari Karonen
19.6k25183
$endgroup$
As a logical proposition, the material conditional $A implies B$ is a very weak one: as you've noticed, it's very easy to satisfy it just by accident. In fact, this happens whenever $A$ is false, or whenever $B$ is true. Thus, merely observing that $A implies B$, for some specific $A$ and $B$, says very little.
Instead, the usefulness of implication lies in the fact that, precisely because of its weakness, it is often possible to assert $A implies B$ as a universal statement (either an axiom or a provable theorem) that holds for any valuation of any free variables mentioned in the propositions $A$ and $B$.
For example, consider the statement: $$x > 2 ;land; x text{ is prime} implies x text{ is odd}.$$ Merely observing that this statement holds for some $x$ says very little — there are plenty of numbers for which it is trivially true, either because they are odd, or because they are not primes greater than 2. What makes this statement useful is that we can prove that it holds for all $x$ — there isn't a single number which would be greater than 2 and prime, but not odd.
answered Sep 27 '14 at 14:23
Ilmari KaronenIlmari Karonen
19.6k25183
edited Mar 30 '15 at 12:10
answered Sep 27 '14 at 14:23
Ilmari KaronenIlmari Karonen
19.6k25183
answered Sep 27 '14 at 14:23
Ilmari KaronenIlmari Karonen
19.6k25183
answered Sep 27 '14 at 14:23
Ilmari KaronenIlmari Karonen
19.6k25183
19.6k25183
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This is the answer that gets to the heart of the matter. +1.
$endgroup$
– 6005
Sep 27 '14 at 16:18
add a comment |
$begingroup$
This is the answer that gets to the heart of the matter. +1.
$endgroup$
– 6005
Sep 27 '14 at 16:18
$begingroup$
This is the answer that gets to the heart of the matter. +1.
$endgroup$
– 6005
Sep 27 '14 at 16:18
$begingroup$
This is the answer that gets to the heart of the matter. +1.
$endgroup$
– 6005
Sep 27 '14 at 16:18
add a comment |
$begingroup$
One way to understand implication is to remember that $ARightarrow B$ is equivalent to $neg A lor B$. If you understand negation ($neg$) and disjunction ($lor$), then you understand implication.
answered Sep 27 '14 at 20:30
Theodore NorvellTheodore Norvell
39119
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7
$begingroup$
I thought this was a strange equivalence, until I noticed myself saying things like: "I didn't know that or I would have done this"
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– Justin
Sep 28 '14 at 4:47
1
$begingroup$
I find $neg[Aland neg B]$ easier to work with.
$endgroup$
– Dan Christensen
Sep 29 '14 at 15:39
add a comment |
$begingroup$
One way to understand implication is to remember that $ARightarrow B$ is equivalent to $neg A lor B$. If you understand negation ($neg$) and disjunction ($lor$), then you understand implication.
answered Sep 27 '14 at 20:30
Theodore NorvellTheodore Norvell
39119
$endgroup$
7
$begingroup$
I thought this was a strange equivalence, until I noticed myself saying things like: "I didn't know that or I would have done this"
$endgroup$
– Justin
Sep 28 '14 at 4:47
1
$begingroup$
I find $neg[Aland neg B]$ easier to work with.
$endgroup$
– Dan Christensen
Sep 29 '14 at 15:39
add a comment |
$begingroup$
One way to understand implication is to remember that $ARightarrow B$ is equivalent to $neg A lor B$. If you understand negation ($neg$) and disjunction ($lor$), then you understand implication.
answered Sep 27 '14 at 20:30
Theodore NorvellTheodore Norvell
39119
$endgroup$
One way to understand implication is to remember that $ARightarrow B$ is equivalent to $neg A lor B$. If you understand negation ($neg$) and disjunction ($lor$), then you understand implication.
answered Sep 27 '14 at 20:30
Theodore NorvellTheodore Norvell
39119
answered Sep 27 '14 at 20:30
Theodore NorvellTheodore Norvell
39119
answered Sep 27 '14 at 20:30
Theodore NorvellTheodore Norvell
39119
answered Sep 27 '14 at 20:30
Theodore NorvellTheodore Norvell
39119
39119
7
$begingroup$
I thought this was a strange equivalence, until I noticed myself saying things like: "I didn't know that or I would have done this"
$endgroup$
– Justin
Sep 28 '14 at 4:47
1
$begingroup$
I find $neg[Aland neg B]$ easier to work with.
$endgroup$
– Dan Christensen
Sep 29 '14 at 15:39
add a comment |
7
$begingroup$
I thought this was a strange equivalence, until I noticed myself saying things like: "I didn't know that or I would have done this"
$endgroup$
– Justin
Sep 28 '14 at 4:47
1
$begingroup$
I find $neg[Aland neg B]$ easier to work with.
$endgroup$
– Dan Christensen
Sep 29 '14 at 15:39
7
7
$begingroup$
I thought this was a strange equivalence, until I noticed myself saying things like: "I didn't know that or I would have done this"
$endgroup$
– Justin
Sep 28 '14 at 4:47
$begingroup$
I thought this was a strange equivalence, until I noticed myself saying things like: "I didn't know that or I would have done this"
$endgroup$
– Justin
Sep 28 '14 at 4:47
1
1
$begingroup$
I find $neg[Aland neg B]$ easier to work with.
$endgroup$
– Dan Christensen
Sep 29 '14 at 15:39
$begingroup$
I find $neg[Aland neg B]$ easier to work with.
$endgroup$
– Dan Christensen
Sep 29 '14 at 15:39
add a comment |
$begingroup$
Look at $A$ and $B$ as something that is either false or true. For
example let $A$ be the event that tomorrow is Tuesday and let $B$
be the event that the day after tomorrow is Wednesday.
Look at
$$
Aimplies B
$$
as a promise - if $A$ is true then so is $B$.
In our example, if $A$ is true then indeed so is $B$ and so the
implication $Aimplies B$ is true.
However, now consider $C$ as the statement that tomorrow is Friday,
and I state
$$
Cimplies B
$$
that is - I promise you that if $C$ will happen so will $B$.
Tomorrow is not Friday (at the time of writing), and so $C$ is false,
regardless of if $B$ is false or true - my promise was kept.
Now regarding the terminology of proofing $B$ is both $A,B$ are
true. Note that statements like
$$
text{My cat walks on four}implies1+1=2
$$
is true, since both are true, but what would it mean to prove $B$ from $A$
?
answered Sep 27 '14 at 10:18
BelgiBelgi
14.5k954113
$endgroup$
add a comment |
$begingroup$
Look at $A$ and $B$ as something that is either false or true. For
example let $A$ be the event that tomorrow is Tuesday and let $B$
be the event that the day after tomorrow is Wednesday.
Look at
$$
Aimplies B
$$
as a promise - if $A$ is true then so is $B$.
In our example, if $A$ is true then indeed so is $B$ and so the
implication $Aimplies B$ is true.
However, now consider $C$ as the statement that tomorrow is Friday,
and I state
$$
Cimplies B
$$
that is - I promise you that if $C$ will happen so will $B$.
Tomorrow is not Friday (at the time of writing), and so $C$ is false,
regardless of if $B$ is false or true - my promise was kept.
Now regarding the terminology of proofing $B$ is both $A,B$ are
true. Note that statements like
$$
text{My cat walks on four}implies1+1=2
$$
is true, since both are true, but what would it mean to prove $B$ from $A$
?
answered Sep 27 '14 at 10:18
BelgiBelgi
14.5k954113
$endgroup$
add a comment |
$begingroup$
Look at $A$ and $B$ as something that is either false or true. For
example let $A$ be the event that tomorrow is Tuesday and let $B$
be the event that the day after tomorrow is Wednesday.
Look at
$$
Aimplies B
$$
as a promise - if $A$ is true then so is $B$.
In our example, if $A$ is true then indeed so is $B$ and so the
implication $Aimplies B$ is true.
However, now consider $C$ as the statement that tomorrow is Friday,
and I state
$$
Cimplies B
$$
that is - I promise you that if $C$ will happen so will $B$.
Tomorrow is not Friday (at the time of writing), and so $C$ is false,
regardless of if $B$ is false or true - my promise was kept.
Now regarding the terminology of proofing $B$ is both $A,B$ are
true. Note that statements like
$$
text{My cat walks on four}implies1+1=2
$$
is true, since both are true, but what would it mean to prove $B$ from $A$
?
answered Sep 27 '14 at 10:18
BelgiBelgi
14.5k954113
$endgroup$
Look at $A$ and $B$ as something that is either false or true. For
example let $A$ be the event that tomorrow is Tuesday and let $B$
be the event that the day after tomorrow is Wednesday.
Look at
$$
Aimplies B
$$
as a promise - if $A$ is true then so is $B$.
In our example, if $A$ is true then indeed so is $B$ and so the
implication $Aimplies B$ is true.
However, now consider $C$ as the statement that tomorrow is Friday,
and I state
$$
Cimplies B
$$
that is - I promise you that if $C$ will happen so will $B$.
Tomorrow is not Friday (at the time of writing), and so $C$ is false,
regardless of if $B$ is false or true - my promise was kept.
Now regarding the terminology of proofing $B$ is both $A,B$ are
true. Note that statements like
$$
text{My cat walks on four}implies1+1=2
$$
is true, since both are true, but what would it mean to prove $B$ from $A$
?
answered Sep 27 '14 at 10:18
BelgiBelgi
14.5k954113
edited Sep 27 '14 at 10:24
answered Sep 27 '14 at 10:18
BelgiBelgi
14.5k954113
answered Sep 27 '14 at 10:18
BelgiBelgi
14.5k954113
answered Sep 27 '14 at 10:18
BelgiBelgi
14.5k954113
14.5k954113
add a comment |
add a comment |
$begingroup$
I would say that $A$ being true and $B$ being true does not mean you can always prove (deduce) $B$ from $A$.
Here's an example. A: Alice lives in Atlanta. B: Bob lives in Boston. Even if these are both true, there is no (apparent) relationship. So you can't logically deduce $B$ from $A$ even though $ARightarrow B$ is true in this case.
I guess this means that if you can logically deduce statement $Q$ from statement $P$, then $PRightarrow Q$ is true; but knowing $PRightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption of $P$.
Logical implication is a defined logical connective, so as long as $P$ and $Q$ have truth values (true or false), so does $PRightarrow Q$.
answered Sep 27 '14 at 10:17
paw88789paw88789
29.1k12349
$endgroup$
$begingroup$
You can logically deduce B from A in that example. Let's assume A is true. B is true. QED. Yes, this counts, unless you really meant A implies B and also not not-A implies not-B. But if you meant that, you should have said so.
$endgroup$
– djechlin
Sep 28 '14 at 2:00
$begingroup$
Your 3rd paragraph is incorrect. Suppose $PRightarrow Q$ is given, we construct a proof of $Q$ to $P$: Assume $P$, assume $neg Q$, then by the truth table for $PRightarrow Q$ (or by modes ponens) $Q$. Hence $Q$ and $neg Q$, a contradiction. Hence our assumption $neg Q$ was false.
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– James
Sep 29 '14 at 16:41
$begingroup$
My point is that sometimes two statements, $P$ and $Q$ are unrelated. Now $PRightarrow Q$ can perfectly well be true (provided $P$ is false or $Q$ is true), and yet knowing that $P$ is true does not help us determine (deduce) that $Q$ is true, because the statements are unrelated.
$endgroup$
– paw88789
Sep 29 '14 at 17:43
$begingroup$
@paw88789 But when you know $P$, then to quote you "knowing $PRightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption that $P$." This is wrong, I just gave a deduction.
$endgroup$
– James
Sep 29 '14 at 18:42
add a comment |
$begingroup$
I would say that $A$ being true and $B$ being true does not mean you can always prove (deduce) $B$ from $A$.
Here's an example. A: Alice lives in Atlanta. B: Bob lives in Boston. Even if these are both true, there is no (apparent) relationship. So you can't logically deduce $B$ from $A$ even though $ARightarrow B$ is true in this case.
I guess this means that if you can logically deduce statement $Q$ from statement $P$, then $PRightarrow Q$ is true; but knowing $PRightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption of $P$.
Logical implication is a defined logical connective, so as long as $P$ and $Q$ have truth values (true or false), so does $PRightarrow Q$.
answered Sep 27 '14 at 10:17
paw88789paw88789
29.1k12349
$endgroup$
$begingroup$
You can logically deduce B from A in that example. Let's assume A is true. B is true. QED. Yes, this counts, unless you really meant A implies B and also not not-A implies not-B. But if you meant that, you should have said so.
$endgroup$
– djechlin
Sep 28 '14 at 2:00
$begingroup$
Your 3rd paragraph is incorrect. Suppose $PRightarrow Q$ is given, we construct a proof of $Q$ to $P$: Assume $P$, assume $neg Q$, then by the truth table for $PRightarrow Q$ (or by modes ponens) $Q$. Hence $Q$ and $neg Q$, a contradiction. Hence our assumption $neg Q$ was false.
$endgroup$
– James
Sep 29 '14 at 16:41
$begingroup$
My point is that sometimes two statements, $P$ and $Q$ are unrelated. Now $PRightarrow Q$ can perfectly well be true (provided $P$ is false or $Q$ is true), and yet knowing that $P$ is true does not help us determine (deduce) that $Q$ is true, because the statements are unrelated.
$endgroup$
– paw88789
Sep 29 '14 at 17:43
$begingroup$
@paw88789 But when you know $P$, then to quote you "knowing $PRightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption that $P$." This is wrong, I just gave a deduction.
$endgroup$
– James
Sep 29 '14 at 18:42
add a comment |
$begingroup$
I would say that $A$ being true and $B$ being true does not mean you can always prove (deduce) $B$ from $A$.
Here's an example. A: Alice lives in Atlanta. B: Bob lives in Boston. Even if these are both true, there is no (apparent) relationship. So you can't logically deduce $B$ from $A$ even though $ARightarrow B$ is true in this case.
I guess this means that if you can logically deduce statement $Q$ from statement $P$, then $PRightarrow Q$ is true; but knowing $PRightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption of $P$.
Logical implication is a defined logical connective, so as long as $P$ and $Q$ have truth values (true or false), so does $PRightarrow Q$.
answered Sep 27 '14 at 10:17
paw88789paw88789
29.1k12349
$endgroup$
I would say that $A$ being true and $B$ being true does not mean you can always prove (deduce) $B$ from $A$.
Here's an example. A: Alice lives in Atlanta. B: Bob lives in Boston. Even if these are both true, there is no (apparent) relationship. So you can't logically deduce $B$ from $A$ even though $ARightarrow B$ is true in this case.
I guess this means that if you can logically deduce statement $Q$ from statement $P$, then $PRightarrow Q$ is true; but knowing $PRightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption of $P$.
Logical implication is a defined logical connective, so as long as $P$ and $Q$ have truth values (true or false), so does $PRightarrow Q$.
answered Sep 27 '14 at 10:17
paw88789paw88789
29.1k12349
answered Sep 27 '14 at 10:17
paw88789paw88789
29.1k12349
answered Sep 27 '14 at 10:17
paw88789paw88789
29.1k12349
answered Sep 27 '14 at 10:17
paw88789paw88789
29.1k12349
29.1k12349
$begingroup$
You can logically deduce B from A in that example. Let's assume A is true. B is true. QED. Yes, this counts, unless you really meant A implies B and also not not-A implies not-B. But if you meant that, you should have said so.
$endgroup$
– djechlin
Sep 28 '14 at 2:00
$begingroup$
Your 3rd paragraph is incorrect. Suppose $PRightarrow Q$ is given, we construct a proof of $Q$ to $P$: Assume $P$, assume $neg Q$, then by the truth table for $PRightarrow Q$ (or by modes ponens) $Q$. Hence $Q$ and $neg Q$, a contradiction. Hence our assumption $neg Q$ was false.
$endgroup$
– James
Sep 29 '14 at 16:41
$begingroup$
My point is that sometimes two statements, $P$ and $Q$ are unrelated. Now $PRightarrow Q$ can perfectly well be true (provided $P$ is false or $Q$ is true), and yet knowing that $P$ is true does not help us determine (deduce) that $Q$ is true, because the statements are unrelated.
$endgroup$
– paw88789
Sep 29 '14 at 17:43
$begingroup$
@paw88789 But when you know $P$, then to quote you "knowing $PRightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption that $P$." This is wrong, I just gave a deduction.
$endgroup$
– James
Sep 29 '14 at 18:42
add a comment |
$begingroup$
You can logically deduce B from A in that example. Let's assume A is true. B is true. QED. Yes, this counts, unless you really meant A implies B and also not not-A implies not-B. But if you meant that, you should have said so.
$endgroup$
– djechlin
Sep 28 '14 at 2:00
$begingroup$
Your 3rd paragraph is incorrect. Suppose $PRightarrow Q$ is given, we construct a proof of $Q$ to $P$: Assume $P$, assume $neg Q$, then by the truth table for $PRightarrow Q$ (or by modes ponens) $Q$. Hence $Q$ and $neg Q$, a contradiction. Hence our assumption $neg Q$ was false.
$endgroup$
– James
Sep 29 '14 at 16:41
$begingroup$
My point is that sometimes two statements, $P$ and $Q$ are unrelated. Now $PRightarrow Q$ can perfectly well be true (provided $P$ is false or $Q$ is true), and yet knowing that $P$ is true does not help us determine (deduce) that $Q$ is true, because the statements are unrelated.
$endgroup$
– paw88789
Sep 29 '14 at 17:43
$begingroup$
@paw88789 But when you know $P$, then to quote you "knowing $PRightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption that $P$." This is wrong, I just gave a deduction.
$endgroup$
– James
Sep 29 '14 at 18:42
$begingroup$
You can logically deduce B from A in that example. Let's assume A is true. B is true. QED. Yes, this counts, unless you really meant A implies B and also not not-A implies not-B. But if you meant that, you should have said so.
$endgroup$
– djechlin
Sep 28 '14 at 2:00
$begingroup$
You can logically deduce B from A in that example. Let's assume A is true. B is true. QED. Yes, this counts, unless you really meant A implies B and also not not-A implies not-B. But if you meant that, you should have said so.
$endgroup$
– djechlin
Sep 28 '14 at 2:00
$begingroup$
Your 3rd paragraph is incorrect. Suppose $PRightarrow Q$ is given, we construct a proof of $Q$ to $P$: Assume $P$, assume $neg Q$, then by the truth table for $PRightarrow Q$ (or by modes ponens) $Q$. Hence $Q$ and $neg Q$, a contradiction. Hence our assumption $neg Q$ was false.
$endgroup$
– James
Sep 29 '14 at 16:41
$begingroup$
Your 3rd paragraph is incorrect. Suppose $PRightarrow Q$ is given, we construct a proof of $Q$ to $P$: Assume $P$, assume $neg Q$, then by the truth table for $PRightarrow Q$ (or by modes ponens) $Q$. Hence $Q$ and $neg Q$, a contradiction. Hence our assumption $neg Q$ was false.
$endgroup$
– James
Sep 29 '14 at 16:41
$begingroup$
My point is that sometimes two statements, $P$ and $Q$ are unrelated. Now $PRightarrow Q$ can perfectly well be true (provided $P$ is false or $Q$ is true), and yet knowing that $P$ is true does not help us determine (deduce) that $Q$ is true, because the statements are unrelated.
$endgroup$
– paw88789
Sep 29 '14 at 17:43
$begingroup$
My point is that sometimes two statements, $P$ and $Q$ are unrelated. Now $PRightarrow Q$ can perfectly well be true (provided $P$ is false or $Q$ is true), and yet knowing that $P$ is true does not help us determine (deduce) that $Q$ is true, because the statements are unrelated.
$endgroup$
– paw88789
Sep 29 '14 at 17:43
$begingroup$
@paw88789 But when you know $P$, then to quote you "knowing $PRightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption that $P$." This is wrong, I just gave a deduction.
$endgroup$
– James
Sep 29 '14 at 18:42
$begingroup$
@paw88789 But when you know $P$, then to quote you "knowing $PRightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption that $P$." This is wrong, I just gave a deduction.
$endgroup$
– James
Sep 29 '14 at 18:42
add a comment |
$begingroup$
Maybe it's more clear if we separate the logical operator meaning of implication from its logical statement meaning.
When we use it as a logical operator, we conceive it simply as an entity, that given two logical values (thus true or false), produces a third logical value, using a common defined rule (its truth table).
So it makes perfect sense to say $A implies B = true$ if $A = B = true$, and we don't concern about what actually proposition $A$ and $B$ means, we care exclusively about their logical values.
It's different when we use implication as a logical statement.
In this case we really say something about the meaning of the propositions involved in our statement. So, while proposition $A = My;cat;is;black$ is true, and proposition $B = I;am;hungry$ is true as well, $A implies B$ is not a valid statement.
Such meaning is linked to set theory and formal logic. Using implication in this context means that you can infer $B$ from $A$, in a way called modus ponens.
answered Sep 27 '14 at 11:11
mattecapumattecapu
695618
$endgroup$
add a comment |
$begingroup$
Maybe it's more clear if we separate the logical operator meaning of implication from its logical statement meaning.
When we use it as a logical operator, we conceive it simply as an entity, that given two logical values (thus true or false), produces a third logical value, using a common defined rule (its truth table).
So it makes perfect sense to say $A implies B = true$ if $A = B = true$, and we don't concern about what actually proposition $A$ and $B$ means, we care exclusively about their logical values.
It's different when we use implication as a logical statement.
In this case we really say something about the meaning of the propositions involved in our statement. So, while proposition $A = My;cat;is;black$ is true, and proposition $B = I;am;hungry$ is true as well, $A implies B$ is not a valid statement.
Such meaning is linked to set theory and formal logic. Using implication in this context means that you can infer $B$ from $A$, in a way called modus ponens.
answered Sep 27 '14 at 11:11
mattecapumattecapu
695618
$endgroup$
add a comment |
$begingroup$
Maybe it's more clear if we separate the logical operator meaning of implication from its logical statement meaning.
When we use it as a logical operator, we conceive it simply as an entity, that given two logical values (thus true or false), produces a third logical value, using a common defined rule (its truth table).
So it makes perfect sense to say $A implies B = true$ if $A = B = true$, and we don't concern about what actually proposition $A$ and $B$ means, we care exclusively about their logical values.
It's different when we use implication as a logical statement.
In this case we really say something about the meaning of the propositions involved in our statement. So, while proposition $A = My;cat;is;black$ is true, and proposition $B = I;am;hungry$ is true as well, $A implies B$ is not a valid statement.
Such meaning is linked to set theory and formal logic. Using implication in this context means that you can infer $B$ from $A$, in a way called modus ponens.
answered Sep 27 '14 at 11:11
mattecapumattecapu
695618
$endgroup$
Maybe it's more clear if we separate the logical operator meaning of implication from its logical statement meaning.
When we use it as a logical operator, we conceive it simply as an entity, that given two logical values (thus true or false), produces a third logical value, using a common defined rule (its truth table).
So it makes perfect sense to say $A implies B = true$ if $A = B = true$, and we don't concern about what actually proposition $A$ and $B$ means, we care exclusively about their logical values.
It's different when we use implication as a logical statement.
In this case we really say something about the meaning of the propositions involved in our statement. So, while proposition $A = My;cat;is;black$ is true, and proposition $B = I;am;hungry$ is true as well, $A implies B$ is not a valid statement.
Such meaning is linked to set theory and formal logic. Using implication in this context means that you can infer $B$ from $A$, in a way called modus ponens.
answered Sep 27 '14 at 11:11
mattecapumattecapu
695618
edited Sep 27 '14 at 11:31
answered Sep 27 '14 at 11:11
mattecapumattecapu
695618
answered Sep 27 '14 at 11:11
mattecapumattecapu
695618
answered Sep 27 '14 at 11:11
mattecapumattecapu
695618
695618
add a comment |
add a comment |
$begingroup$
Implication can be understood as thinking in NECESSARY and SUFFICIENT conditions. A good example is to think about someone who has born in Dallas, Texas.
Proposition P could stand for "Someone who was born in Dallas"
Proposition Q could stand for "Someone who is texan"
Truth table for implication is:
So, in this example is easy to check line by line the validity of truth table above:
Who was born in Dallas is therefore texan; So the first implication is TRUE.
Who was born in Dallas MUST be texan. So, saying that someone who was born in Dallas, in the state of Texas, and is not texan is FALSE;
Saying who wasn't born in Dallas but is texan anyway (take someone who has born in Houston, for example) is a TRUE statement;
At last, who wasn't born in Dallas AND also is NOT texan is also TRUE.
So, thinking in terms of NECESSARY (to be texan is a necessary condition for who was born in Dallas) and SUFFICIENT (to born in Dallas is a sufficient condition to be texan) conditions ease the process to translate implication to natural language.
Other examples of propositions containing necessary and sufficient conditions are:
- P = To be a priest / Q = To believe in God;
- P = To be a Police officer / Q = To carry a gun;
- P = The number is divisible by 4 / Q = The number is even.
answered Mar 30 '15 at 9:37
Daniel LavieriDaniel Lavieri
213
$endgroup$
add a comment |
$begingroup$
Implication can be understood as thinking in NECESSARY and SUFFICIENT conditions. A good example is to think about someone who has born in Dallas, Texas.
Proposition P could stand for "Someone who was born in Dallas"
Proposition Q could stand for "Someone who is texan"
Truth table for implication is:
So, in this example is easy to check line by line the validity of truth table above:
Who was born in Dallas is therefore texan; So the first implication is TRUE.
Who was born in Dallas MUST be texan. So, saying that someone who was born in Dallas, in the state of Texas, and is not texan is FALSE;
Saying who wasn't born in Dallas but is texan anyway (take someone who has born in Houston, for example) is a TRUE statement;
At last, who wasn't born in Dallas AND also is NOT texan is also TRUE.
So, thinking in terms of NECESSARY (to be texan is a necessary condition for who was born in Dallas) and SUFFICIENT (to born in Dallas is a sufficient condition to be texan) conditions ease the process to translate implication to natural language.
Other examples of propositions containing necessary and sufficient conditions are:
- P = To be a priest / Q = To believe in God;
- P = To be a Police officer / Q = To carry a gun;
- P = The number is divisible by 4 / Q = The number is even.
answered Mar 30 '15 at 9:37
Daniel LavieriDaniel Lavieri
213
$endgroup$
add a comment |
$begingroup$
Implication can be understood as thinking in NECESSARY and SUFFICIENT conditions. A good example is to think about someone who has born in Dallas, Texas.
Proposition P could stand for "Someone who was born in Dallas"
Proposition Q could stand for "Someone who is texan"
Truth table for implication is:
So, in this example is easy to check line by line the validity of truth table above:
Who was born in Dallas is therefore texan; So the first implication is TRUE.
Who was born in Dallas MUST be texan. So, saying that someone who was born in Dallas, in the state of Texas, and is not texan is FALSE;
Saying who wasn't born in Dallas but is texan anyway (take someone who has born in Houston, for example) is a TRUE statement;
At last, who wasn't born in Dallas AND also is NOT texan is also TRUE.
So, thinking in terms of NECESSARY (to be texan is a necessary condition for who was born in Dallas) and SUFFICIENT (to born in Dallas is a sufficient condition to be texan) conditions ease the process to translate implication to natural language.
Other examples of propositions containing necessary and sufficient conditions are:
- P = To be a priest / Q = To believe in God;
- P = To be a Police officer / Q = To carry a gun;
- P = The number is divisible by 4 / Q = The number is even.
answered Mar 30 '15 at 9:37
Daniel LavieriDaniel Lavieri
213
$endgroup$
Implication can be understood as thinking in NECESSARY and SUFFICIENT conditions. A good example is to think about someone who has born in Dallas, Texas.
Proposition P could stand for "Someone who was born in Dallas"
Proposition Q could stand for "Someone who is texan"
Truth table for implication is:
So, in this example is easy to check line by line the validity of truth table above:
Who was born in Dallas is therefore texan; So the first implication is TRUE.
Who was born in Dallas MUST be texan. So, saying that someone who was born in Dallas, in the state of Texas, and is not texan is FALSE;
Saying who wasn't born in Dallas but is texan anyway (take someone who has born in Houston, for example) is a TRUE statement;
At last, who wasn't born in Dallas AND also is NOT texan is also TRUE.
So, thinking in terms of NECESSARY (to be texan is a necessary condition for who was born in Dallas) and SUFFICIENT (to born in Dallas is a sufficient condition to be texan) conditions ease the process to translate implication to natural language.
Other examples of propositions containing necessary and sufficient conditions are:
- P = To be a priest / Q = To believe in God;
- P = To be a Police officer / Q = To carry a gun;
- P = The number is divisible by 4 / Q = The number is even.
answered Mar 30 '15 at 9:37
Daniel LavieriDaniel Lavieri
213
edited Mar 30 '15 at 16:19
answered Mar 30 '15 at 9:37
Daniel LavieriDaniel Lavieri
213
answered Mar 30 '15 at 9:37
Daniel LavieriDaniel Lavieri
213
answered Mar 30 '15 at 9:37
Daniel LavieriDaniel Lavieri
213
213
add a comment |
add a comment |
$begingroup$
The problem is understanding that there's a difference between the everyday mathematical meaning of "implies" - "I can prove it (or someone can)" - and the formal logical meaning expressed by the truth table. This may help: In classical logic, why is $(pRightarrow q)$ True if $p$ is False and $q$ is True?.
edited Apr 13 '17 at 12:21
Community♦
1
answered Sep 27 '14 at 22:22
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
add a comment |
$begingroup$
The problem is understanding that there's a difference between the everyday mathematical meaning of "implies" - "I can prove it (or someone can)" - and the formal logical meaning expressed by the truth table. This may help: In classical logic, why is $(pRightarrow q)$ True if $p$ is False and $q$ is True?.
edited Apr 13 '17 at 12:21
Community♦
1
answered Sep 27 '14 at 22:22
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
add a comment |
$begingroup$
The problem is understanding that there's a difference between the everyday mathematical meaning of "implies" - "I can prove it (or someone can)" - and the formal logical meaning expressed by the truth table. This may help: In classical logic, why is $(pRightarrow q)$ True if $p$ is False and $q$ is True?.
edited Apr 13 '17 at 12:21
Community♦
1
answered Sep 27 '14 at 22:22
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
The problem is understanding that there's a difference between the everyday mathematical meaning of "implies" - "I can prove it (or someone can)" - and the formal logical meaning expressed by the truth table. This may help: In classical logic, why is $(pRightarrow q)$ True if $p$ is False and $q$ is True?.
edited Apr 13 '17 at 12:21
Community♦
1
answered Sep 27 '14 at 22:22
Ethan BolkerEthan Bolker
42.1k548111
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Sep 27 '14 at 22:22
Ethan BolkerEthan Bolker
42.1k548111
answered Sep 27 '14 at 22:22
Ethan BolkerEthan Bolker
42.1k548111
answered Sep 27 '14 at 22:22
Ethan BolkerEthan Bolker
42.1k548111
42.1k548111
add a comment |
add a comment |
$begingroup$
If sentence $A$ is true and $B$ is true then $Aimplies B$ is true.
Yes. In mathematics and logic as widely applied, $Aimplies B$ is equivalent to $neg[Alandneg B]$.
So, if $A$ is true and $B$ is true, then $neg[Alandneg B]$ and hence $Aimplies B$ will also be true.
Does that mean if $A$ and $B$ are both true then there is a way to prove $B$ is true from $A$, always?
Yes. Here is an informal proof:
Suppose $A land B$.
Suppose $A$.
$B$ (from 1)
$Aimplies B$ (conclusion from 2 and 3)
$Aland B implies [Aimplies B]$ (conclusion from 1 and 4)
Alternatively, prove (5) using truth table.
The same for if $A$ is false can you get anything either True or false proved from this $A$?
Yes. Informal proof:
Suppose $neg A$
Suppose $Aland neg B$
$Aland neg A$ (contradiction from 1 and 2)
$neg[Aland neg B]$ (by contradiciton from 2 and 3)
$Aimplies B$ from (equivalently from 4)
$neg Aimplies [Aimplies B] $ (conclusion from 1 and 5)
Alternatively, prove (6) using a truth table.
That the above may seem counter-intuitive may stem from the fact that implication is often confused with temporal notions of causality. The statement, "If it is raining, then it is cloudy" on its own does not mean that rain always causes clouds, or that clouds always cause rain. If means only that, at a given instant in time, its is not the case that it is raining and not cloudy.
answered Sep 29 '14 at 15:55
Dan ChristensenDan Christensen
8,55321833
$endgroup$
add a comment |
$begingroup$
If sentence $A$ is true and $B$ is true then $Aimplies B$ is true.
Yes. In mathematics and logic as widely applied, $Aimplies B$ is equivalent to $neg[Alandneg B]$.
So, if $A$ is true and $B$ is true, then $neg[Alandneg B]$ and hence $Aimplies B$ will also be true.
Does that mean if $A$ and $B$ are both true then there is a way to prove $B$ is true from $A$, always?
Yes. Here is an informal proof:
Suppose $A land B$.
Suppose $A$.
$B$ (from 1)
$Aimplies B$ (conclusion from 2 and 3)
$Aland B implies [Aimplies B]$ (conclusion from 1 and 4)
Alternatively, prove (5) using truth table.
The same for if $A$ is false can you get anything either True or false proved from this $A$?
Yes. Informal proof:
Suppose $neg A$
Suppose $Aland neg B$
$Aland neg A$ (contradiction from 1 and 2)
$neg[Aland neg B]$ (by contradiciton from 2 and 3)
$Aimplies B$ from (equivalently from 4)
$neg Aimplies [Aimplies B] $ (conclusion from 1 and 5)
Alternatively, prove (6) using a truth table.
That the above may seem counter-intuitive may stem from the fact that implication is often confused with temporal notions of causality. The statement, "If it is raining, then it is cloudy" on its own does not mean that rain always causes clouds, or that clouds always cause rain. If means only that, at a given instant in time, its is not the case that it is raining and not cloudy.
answered Sep 29 '14 at 15:55
Dan ChristensenDan Christensen
8,55321833
$endgroup$
add a comment |
$begingroup$
If sentence $A$ is true and $B$ is true then $Aimplies B$ is true.
Yes. In mathematics and logic as widely applied, $Aimplies B$ is equivalent to $neg[Alandneg B]$.
So, if $A$ is true and $B$ is true, then $neg[Alandneg B]$ and hence $Aimplies B$ will also be true.
Does that mean if $A$ and $B$ are both true then there is a way to prove $B$ is true from $A$, always?
Yes. Here is an informal proof:
Suppose $A land B$.
Suppose $A$.
$B$ (from 1)
$Aimplies B$ (conclusion from 2 and 3)
$Aland B implies [Aimplies B]$ (conclusion from 1 and 4)
Alternatively, prove (5) using truth table.
The same for if $A$ is false can you get anything either True or false proved from this $A$?
Yes. Informal proof:
Suppose $neg A$
Suppose $Aland neg B$
$Aland neg A$ (contradiction from 1 and 2)
$neg[Aland neg B]$ (by contradiciton from 2 and 3)
$Aimplies B$ from (equivalently from 4)
$neg Aimplies [Aimplies B] $ (conclusion from 1 and 5)
Alternatively, prove (6) using a truth table.
That the above may seem counter-intuitive may stem from the fact that implication is often confused with temporal notions of causality. The statement, "If it is raining, then it is cloudy" on its own does not mean that rain always causes clouds, or that clouds always cause rain. If means only that, at a given instant in time, its is not the case that it is raining and not cloudy.
answered Sep 29 '14 at 15:55
Dan ChristensenDan Christensen
8,55321833
$endgroup$
If sentence $A$ is true and $B$ is true then $Aimplies B$ is true.
Yes. In mathematics and logic as widely applied, $Aimplies B$ is equivalent to $neg[Alandneg B]$.
So, if $A$ is true and $B$ is true, then $neg[Alandneg B]$ and hence $Aimplies B$ will also be true.
Does that mean if $A$ and $B$ are both true then there is a way to prove $B$ is true from $A$, always?
Yes. Here is an informal proof:
Suppose $A land B$.
Suppose $A$.
$B$ (from 1)
$Aimplies B$ (conclusion from 2 and 3)
$Aland B implies [Aimplies B]$ (conclusion from 1 and 4)
Alternatively, prove (5) using truth table.
The same for if $A$ is false can you get anything either True or false proved from this $A$?
Yes. Informal proof:
Suppose $neg A$
Suppose $Aland neg B$
$Aland neg A$ (contradiction from 1 and 2)
$neg[Aland neg B]$ (by contradiciton from 2 and 3)
$Aimplies B$ from (equivalently from 4)
$neg Aimplies [Aimplies B] $ (conclusion from 1 and 5)
Alternatively, prove (6) using a truth table.
That the above may seem counter-intuitive may stem from the fact that implication is often confused with temporal notions of causality. The statement, "If it is raining, then it is cloudy" on its own does not mean that rain always causes clouds, or that clouds always cause rain. If means only that, at a given instant in time, its is not the case that it is raining and not cloudy.
answered Sep 29 '14 at 15:55
Dan ChristensenDan Christensen
8,55321833
edited Sep 29 '14 at 18:49
answered Sep 29 '14 at 15:55
Dan ChristensenDan Christensen
8,55321833
answered Sep 29 '14 at 15:55
Dan ChristensenDan Christensen
8,55321833
answered Sep 29 '14 at 15:55
Dan ChristensenDan Christensen
8,55321833
8,55321833
add a comment |
add a comment |
$begingroup$
The sentence $Aimplies B$ is not a sentence about proving anything. It is a compound sentence, but otherwise just like the sentences $A$ and $B$. For the truth table, the compound sentence got to have truth values like $A$ and $B$ has, and it has to depend on these sentences truth values. The values for this compound sentence is the same as for the compound statement $neg A vee B$.
answered Sep 29 '14 at 19:14
LehsLehs
6,98031662
$endgroup$
add a comment |
$begingroup$
The sentence $Aimplies B$ is not a sentence about proving anything. It is a compound sentence, but otherwise just like the sentences $A$ and $B$. For the truth table, the compound sentence got to have truth values like $A$ and $B$ has, and it has to depend on these sentences truth values. The values for this compound sentence is the same as for the compound statement $neg A vee B$.
answered Sep 29 '14 at 19:14
LehsLehs
6,98031662
$endgroup$
add a comment |
$begingroup$
The sentence $Aimplies B$ is not a sentence about proving anything. It is a compound sentence, but otherwise just like the sentences $A$ and $B$. For the truth table, the compound sentence got to have truth values like $A$ and $B$ has, and it has to depend on these sentences truth values. The values for this compound sentence is the same as for the compound statement $neg A vee B$.
answered Sep 29 '14 at 19:14
LehsLehs
6,98031662
$endgroup$
The sentence $Aimplies B$ is not a sentence about proving anything. It is a compound sentence, but otherwise just like the sentences $A$ and $B$. For the truth table, the compound sentence got to have truth values like $A$ and $B$ has, and it has to depend on these sentences truth values. The values for this compound sentence is the same as for the compound statement $neg A vee B$.
answered Sep 29 '14 at 19:14
LehsLehs
6,98031662
answered Sep 29 '14 at 19:14
LehsLehs
6,98031662
answered Sep 29 '14 at 19:14
LehsLehs
6,98031662
answered Sep 29 '14 at 19:14
LehsLehs
6,98031662
6,98031662
add a comment |
add a comment |
$begingroup$
This is the way I understand it (I make no guarantee this is considered a correct interpretation).
the truth value of an implies statement is basically 'innocent until proven guilty', or 'my statement is true so long as the proposition values don't disprove it'.
So with an example, propositions P and Q, with the statement P implies Q. Lets say proposition P is 'is a man' and proposition Q is 'lives in japan', so our implication statement is basically saying 'all men live in japan'.
The truth table:
is basically saying, for any example where P and Q are true, we haven't disproved our statement that P -> Q, and for any example where P is false we haven't disproved our statement, only if we have an example of P being true and Q being false is our statement shown to be false.
So imagine there are only 4 humans in existence, we look at each of them and categorise them in our truth table. By doing this we can work out if our statement that P -> Q is true. We find 2 of the people are men, and that they do live in Japan, OK good, our statement still holds true, the 3rd person is a woman, we don't care if she lives in Japan or not because in either case our statement about men is still holding true.
A woman living in or outside japan has no bearing on the truth of our statement about men living in japan.
The 4th person then will decide if our statement is true or not, if the 4th person is a man living in japan, or a woman, then is it logically true to say that being a man implies that you live in japan.
Why is this useful? well the truth table itself doesn't help us prove the truth of our statement by itself, what it does is clearly set out what conditions are relevant to the truth of our statement - i.e. that we don't care about women, and that only a man not living in japan will falsify our statement.
Of course, you could make a statement like 'all green men live on mars'. This is a true statement because we cannot verify who lives on mars, and we don't have any examples of green men. But this is whats called a "vacuous truth", it has no bearing on reality for us because the propositions are intangible.
answered Nov 22 '17 at 9:13
Coder375Coder375
101
$endgroup$
add a comment |
$begingroup$
This is the way I understand it (I make no guarantee this is considered a correct interpretation).
the truth value of an implies statement is basically 'innocent until proven guilty', or 'my statement is true so long as the proposition values don't disprove it'.
So with an example, propositions P and Q, with the statement P implies Q. Lets say proposition P is 'is a man' and proposition Q is 'lives in japan', so our implication statement is basically saying 'all men live in japan'.
The truth table:
is basically saying, for any example where P and Q are true, we haven't disproved our statement that P -> Q, and for any example where P is false we haven't disproved our statement, only if we have an example of P being true and Q being false is our statement shown to be false.
So imagine there are only 4 humans in existence, we look at each of them and categorise them in our truth table. By doing this we can work out if our statement that P -> Q is true. We find 2 of the people are men, and that they do live in Japan, OK good, our statement still holds true, the 3rd person is a woman, we don't care if she lives in Japan or not because in either case our statement about men is still holding true.
A woman living in or outside japan has no bearing on the truth of our statement about men living in japan.
The 4th person then will decide if our statement is true or not, if the 4th person is a man living in japan, or a woman, then is it logically true to say that being a man implies that you live in japan.
Why is this useful? well the truth table itself doesn't help us prove the truth of our statement by itself, what it does is clearly set out what conditions are relevant to the truth of our statement - i.e. that we don't care about women, and that only a man not living in japan will falsify our statement.
Of course, you could make a statement like 'all green men live on mars'. This is a true statement because we cannot verify who lives on mars, and we don't have any examples of green men. But this is whats called a "vacuous truth", it has no bearing on reality for us because the propositions are intangible.
answered Nov 22 '17 at 9:13
Coder375Coder375
101
$endgroup$
add a comment |
$begingroup$
This is the way I understand it (I make no guarantee this is considered a correct interpretation).
the truth value of an implies statement is basically 'innocent until proven guilty', or 'my statement is true so long as the proposition values don't disprove it'.
So with an example, propositions P and Q, with the statement P implies Q. Lets say proposition P is 'is a man' and proposition Q is 'lives in japan', so our implication statement is basically saying 'all men live in japan'.
The truth table:
is basically saying, for any example where P and Q are true, we haven't disproved our statement that P -> Q, and for any example where P is false we haven't disproved our statement, only if we have an example of P being true and Q being false is our statement shown to be false.
So imagine there are only 4 humans in existence, we look at each of them and categorise them in our truth table. By doing this we can work out if our statement that P -> Q is true. We find 2 of the people are men, and that they do live in Japan, OK good, our statement still holds true, the 3rd person is a woman, we don't care if she lives in Japan or not because in either case our statement about men is still holding true.
A woman living in or outside japan has no bearing on the truth of our statement about men living in japan.
The 4th person then will decide if our statement is true or not, if the 4th person is a man living in japan, or a woman, then is it logically true to say that being a man implies that you live in japan.
Why is this useful? well the truth table itself doesn't help us prove the truth of our statement by itself, what it does is clearly set out what conditions are relevant to the truth of our statement - i.e. that we don't care about women, and that only a man not living in japan will falsify our statement.
Of course, you could make a statement like 'all green men live on mars'. This is a true statement because we cannot verify who lives on mars, and we don't have any examples of green men. But this is whats called a "vacuous truth", it has no bearing on reality for us because the propositions are intangible.
answered Nov 22 '17 at 9:13
Coder375Coder375
101
$endgroup$
This is the way I understand it (I make no guarantee this is considered a correct interpretation).
the truth value of an implies statement is basically 'innocent until proven guilty', or 'my statement is true so long as the proposition values don't disprove it'.
So with an example, propositions P and Q, with the statement P implies Q. Lets say proposition P is 'is a man' and proposition Q is 'lives in japan', so our implication statement is basically saying 'all men live in japan'.
The truth table:
is basically saying, for any example where P and Q are true, we haven't disproved our statement that P -> Q, and for any example where P is false we haven't disproved our statement, only if we have an example of P being true and Q being false is our statement shown to be false.
So imagine there are only 4 humans in existence, we look at each of them and categorise them in our truth table. By doing this we can work out if our statement that P -> Q is true. We find 2 of the people are men, and that they do live in Japan, OK good, our statement still holds true, the 3rd person is a woman, we don't care if she lives in Japan or not because in either case our statement about men is still holding true.
A woman living in or outside japan has no bearing on the truth of our statement about men living in japan.
The 4th person then will decide if our statement is true or not, if the 4th person is a man living in japan, or a woman, then is it logically true to say that being a man implies that you live in japan.
Why is this useful? well the truth table itself doesn't help us prove the truth of our statement by itself, what it does is clearly set out what conditions are relevant to the truth of our statement - i.e. that we don't care about women, and that only a man not living in japan will falsify our statement.
Of course, you could make a statement like 'all green men live on mars'. This is a true statement because we cannot verify who lives on mars, and we don't have any examples of green men. But this is whats called a "vacuous truth", it has no bearing on reality for us because the propositions are intangible.
answered Nov 22 '17 at 9:13
Coder375Coder375
101
answered Nov 22 '17 at 9:13
Coder375Coder375
101
answered Nov 22 '17 at 9:13
Coder375Coder375
101
answered Nov 22 '17 at 9:13
Coder375Coder375
101
101
add a comment |
add a comment |
$begingroup$
This illustration of logical implication might help:
a) When you truly understand logical implication,
b) then you’ll be a happy person.
You may be a happy person for other reasons. But if you’re unhappy, then certainly you don’t truly understand logical implication. :-)
Truth table:
a b a⇒b
F F T
F T T
T F F
T T T
I hope this somewhat intuitive example helps.
answered Jan 6 at 5:52
Bert SierraBert Sierra
12
$endgroup$
add a comment |
$begingroup$
This illustration of logical implication might help:
a) When you truly understand logical implication,
b) then you’ll be a happy person.
You may be a happy person for other reasons. But if you’re unhappy, then certainly you don’t truly understand logical implication. :-)
Truth table:
a b a⇒b
F F T
F T T
T F F
T T T
I hope this somewhat intuitive example helps.
answered Jan 6 at 5:52
Bert SierraBert Sierra
12
$endgroup$
add a comment |
$begingroup$
This illustration of logical implication might help:
a) When you truly understand logical implication,
b) then you’ll be a happy person.
You may be a happy person for other reasons. But if you’re unhappy, then certainly you don’t truly understand logical implication. :-)
Truth table:
a b a⇒b
F F T
F T T
T F F
T T T
I hope this somewhat intuitive example helps.
answered Jan 6 at 5:52
Bert SierraBert Sierra
12
$endgroup$
This illustration of logical implication might help:
a) When you truly understand logical implication,
b) then you’ll be a happy person.
You may be a happy person for other reasons. But if you’re unhappy, then certainly you don’t truly understand logical implication. :-)
Truth table:
a b a⇒b
F F T
F T T
T F F
T T T
I hope this somewhat intuitive example helps.
answered Jan 6 at 5:52
Bert SierraBert Sierra
12
edited Jan 6 at 9:41
answered Jan 6 at 5:52
Bert SierraBert Sierra
12
answered Jan 6 at 5:52
Bert SierraBert Sierra
12
answered Jan 6 at 5:52
Bert SierraBert Sierra
12
12
add a comment |
add a comment |
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10
$begingroup$
@gebruiker Yes, you can.
$endgroup$
– Git Gud
Sep 27 '14 at 10:10
3
$begingroup$
@gebruiker - Why would you think you can't ?
$endgroup$
– Belgi
Sep 27 '14 at 10:20
3
$begingroup$
@Belgi Because I own a dog is true, and I am a man is true. But owning a dog does not imply being a man. I know women who own dogs too. Or am I overlooking something here?
$endgroup$
– gebruiker
Sep 27 '14 at 10:32
10
$begingroup$
@gebruiker "(gebruiker owns a dog) implies (gebruiker is a man)" is true. "(X owns a dog) implies (X is a man)" is not true for all X
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– Ben Aaronson
Sep 27 '14 at 20:33
6
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$Aimplies B ;;;;$ means $;;; operatorname{truthvalue}(A) leq operatorname{truthvalue}(B) ;;;$. $;;;;;;;$
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– user57159
Sep 28 '14 at 2:01