Show the logical equivalence of $p to q to perp$ and $p wedge q to perp$. [on hold]

-1

I'm asked to show the logical equivalence of $p to q to perp$ and $p wedge q to perp$. When I make the truth tables, the two statements seem to come out to two very different outcomes, perhaps I'm interpreting the question wrong and the answer is that they aren't logically equivalent?

edited yesterday

metamorphy

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asked yesterday

Richard Cameron

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Which values of $p$ and $q$ did you take to get the "two very different outcomes"?
– 5xum
yesterday

3

Where are parentheses ? In order to anwer correctly, you have to put them...
– Mauro ALLEGRANZA
yesterday

We have that $lnot (p to q)$ is equivalent to $(p land lnot q)$
– Mauro ALLEGRANZA
yesterday

And also that $(p to lnot q)$ is equiv to $lnot (p land q)$.
– Mauro ALLEGRANZA
yesterday

You need brackets to identify what goes with what. When you are combining binary operators which need not be associative, this is crucial.
– Mark Bennet
yesterday

add a comment |

-1

edited yesterday

metamorphy

3,5721521

asked yesterday

Richard Cameron

New contributor

put on hold as off-topic by Mauro ALLEGRANZA, amWhy, José Carlos Santos, KReiser, Shailesh yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, José Carlos Santos, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.

Which values of $p$ and $q$ did you take to get the "two very different outcomes"?
– 5xum
yesterday

3

Where are parentheses ? In order to anwer correctly, you have to put them...
– Mauro ALLEGRANZA
yesterday

We have that $lnot (p to q)$ is equivalent to $(p land lnot q)$
– Mauro ALLEGRANZA
yesterday

And also that $(p to lnot q)$ is equiv to $lnot (p land q)$.
– Mauro ALLEGRANZA
yesterday

You need brackets to identify what goes with what. When you are combining binary operators which need not be associative, this is crucial.
– Mark Bennet
yesterday

add a comment |

-1

edited yesterday

metamorphy

3,5721521

asked yesterday

Richard Cameron

New contributor

propositional-calculus

edited yesterday

metamorphy

3,5721521

asked yesterday

Richard Cameron

New contributor

edited yesterday

metamorphy

3,5721521

asked yesterday

Richard Cameron

New contributor

edited yesterday

metamorphy

3,5721521

edited yesterday

metamorphy

3,5721521

edited yesterday

metamorphy

3,5721521

asked yesterday

Richard Cameron

New contributor

asked yesterday

Richard Cameron

asked yesterday

Richard Cameron

New contributor

Richard Cameron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Mauro ALLEGRANZA, amWhy, José Carlos Santos, KReiser, Shailesh yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, José Carlos Santos, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by Mauro ALLEGRANZA, amWhy, José Carlos Santos, KReiser, Shailesh yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, José Carlos Santos, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.

Which values of $p$ and $q$ did you take to get the "two very different outcomes"?
– 5xum
yesterday

3

Where are parentheses ? In order to anwer correctly, you have to put them...
– Mauro ALLEGRANZA
yesterday

We have that $lnot (p to q)$ is equivalent to $(p land lnot q)$
– Mauro ALLEGRANZA
yesterday

And also that $(p to lnot q)$ is equiv to $lnot (p land q)$.
– Mauro ALLEGRANZA
yesterday

You need brackets to identify what goes with what. When you are combining binary operators which need not be associative, this is crucial.
– Mark Bennet
yesterday

add a comment |

Which values of $p$ and $q$ did you take to get the "two very different outcomes"?
– 5xum
yesterday

3

Where are parentheses ? In order to anwer correctly, you have to put them...
– Mauro ALLEGRANZA
yesterday

We have that $lnot (p to q)$ is equivalent to $(p land lnot q)$
– Mauro ALLEGRANZA
yesterday

And also that $(p to lnot q)$ is equiv to $lnot (p land q)$.
– Mauro ALLEGRANZA
yesterday

You need brackets to identify what goes with what. When you are combining binary operators which need not be associative, this is crucial.
– Mark Bennet
yesterday

Which values of $p$ and $q$ did you take to get the "two very different outcomes"?
– 5xum
yesterday

Where are parentheses ? In order to anwer correctly, you have to put them...
– Mauro ALLEGRANZA
yesterday

We have that $lnot (p to q)$ is equivalent to $(p land lnot q)$
– Mauro ALLEGRANZA
yesterday

And also that $(p to lnot q)$ is equiv to $lnot (p land q)$.
– Mauro ALLEGRANZA
yesterday

You need brackets to identify what goes with what. When you are combining binary operators which need not be associative, this is crucial.
– Mark Bennet
yesterday

add a comment |

1 Answer
1

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First, note that $x to perp = lnot x lor perp = lnot x$. So

$p wedge q to perp = lnot(p land q)$

$(p to q) to perp = lnot(p to q) = lnot(lnot p lor q) = p land lnot q$

$p to (q to perp) = p to lnot q = lnot p lor lnot q = lnot (p land q)$

So, as pointed out in the comments above, it all depends where you put the parentheses when you interpret $p to q to perp$.

answered yesterday

gandalf61

7,703623

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

First, note that $x to perp = lnot x lor perp = lnot x$. So

$p wedge q to perp = lnot(p land q)$

$(p to q) to perp = lnot(p to q) = lnot(lnot p lor q) = p land lnot q$

$p to (q to perp) = p to lnot q = lnot p lor lnot q = lnot (p land q)$

So, as pointed out in the comments above, it all depends where you put the parentheses when you interpret $p to q to perp$.

answered yesterday

gandalf61

7,703623

add a comment |

First, note that $x to perp = lnot x lor perp = lnot x$. So

$p wedge q to perp = lnot(p land q)$

$(p to q) to perp = lnot(p to q) = lnot(lnot p lor q) = p land lnot q$

$p to (q to perp) = p to lnot q = lnot p lor lnot q = lnot (p land q)$

So, as pointed out in the comments above, it all depends where you put the parentheses when you interpret $p to q to perp$.

answered yesterday

gandalf61

7,703623

add a comment |

First, note that $x to perp = lnot x lor perp = lnot x$. So

$p wedge q to perp = lnot(p land q)$

$(p to q) to perp = lnot(p to q) = lnot(lnot p lor q) = p land lnot q$

$p to (q to perp) = p to lnot q = lnot p lor lnot q = lnot (p land q)$

So, as pointed out in the comments above, it all depends where you put the parentheses when you interpret $p to q to perp$.

answered yesterday

gandalf61

7,703623

First, note that $x to perp = lnot x lor perp = lnot x$. So

$p wedge q to perp = lnot(p land q)$

$(p to q) to perp = lnot(p to q) = lnot(lnot p lor q) = p land lnot q$

$p to (q to perp) = p to lnot q = lnot p lor lnot q = lnot (p land q)$

So, as pointed out in the comments above, it all depends where you put the parentheses when you interpret $p to q to perp$.

answered yesterday

gandalf61

7,703623

answered yesterday

gandalf61

7,703623

answered yesterday

gandalf61

7,703623

answered yesterday

gandalf61

7,703623

add a comment |

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