Logical equivalence help Watch

khanpatel321
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#1
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I have the statement p ∨ q.

1) p∧∼q ≡ p ∨ q

I got no it is not.. I used the truth tables with heading as below

p q ~p ~p -> q p ∧ q

My last two columns were not the same, but in the mark scheme it says they are logically equivalent?
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ghostwalker
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(Original post by khanpatel321)
I have the statement p ∨ q.

1) p∧∼q ≡ p ∨ q

I got no it is not.
That's correct, they're not equivalent.
There are two combinations on the left which evaluate as false, whereas there's only one combination on the right.

. I used the truth tables with heading as below

p q ~p ~p -> q p ∧ q

My last two columns were not the same, but in the mark scheme it says they are logically equivalent?
Those column headings don't make sense to me in relation to this question.

Do you have the question and your headings down correctly?
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Doctor_Einstein
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(Original post by khanpatel321)
I have the statement p ∨ q.

1) p∧∼q ≡ p ∨ q

I got no it is not.. I used the truth tables with heading as below

p q ~p ~p -> q p ∧ q

My last two columns were not the same, but in the mark scheme it says they are logically equivalent?
You don't even need a truth table. p ∨ q means you have p or q. However p∧∼q means you have p and you don't have q. Clearly the latter says it is impossible to have q, where as the former says you may have q or p. Therefore they are not equivalent.

So while truth tables are useful to show working, I think you should practice working it out by pure reasoning as well.
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khanpatel321
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(Original post by ghostwalker)
That's correct, they're not equivalent.
There are two combinations on the left which evaluate as false, whereas there's only one combination on the right.



Those column headings don't make sense to me in relation to this question.

Do you have the question and your headings down correctly?
(Original post by Doctor_Einstein)
You don't even need a truth table. p ∨ q means you have p or q. However p∧∼q means you have p and you don't have q. Clearly the latter says it is impossible to have q, where as the former says you may have q or p. Therefore they are not equivalent.

So while truth tables are useful to show working, I think you should practice working it out by pure reasoning as well.
This is the question the last part


This is the answer


So they are definitely not equivalent and the answer in the mark schemes is wrong?

I used truth tables and got that they are not.
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ghostwalker
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(Original post by khanpatel321)
This is the answer


So they are definitely not equivalent and the answer in the mark schemes is wrong?

I used truth tables and got that they are not.
Yes they are equivalent.

But this isn't what you asked in your first post.

Edit: I'd have another go at that truth table.
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davros
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(Original post by khanpatel321)
This is the question the last part


This is the answer


So they are definitely not equivalent and the answer in the mark schemes is wrong?

I used truth tables and got that they are not.
What are you querying exactly?

p -> q is equivalent to ~p V q

So ~p -> q is equivalent to p V q which is the form originally given in the question. So as far as I can see, the answer (vi) is correct. (I haven't bothered eliminating all the other possibilities).
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ghostwalker
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(Original post by davros)
(I haven't bothered eliminating all the other possibilities).
There is more than one that is logically equivalent.
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davros
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(Original post by ghostwalker)
There is more than one that is logically equivalent.
That's fair enough - I was assuming the OP was disagreeing with the answer (vi) but as you said before he hasn't posted consistently what he's comparing!
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khanpatel321
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(Original post by ghostwalker)
Yes they are equivalent.

But this isn't what you asked in your first post.
(Original post by davros)
What are you querying exactly?

p -> q is equivalent to ~p V q

So ~p -> q is equivalent to p V q which is the form originally given in the question. So as far as I can see, the answer (vi) is correct. (I haven't bothered eliminating all the other possibilities).
I'm asking this

is p q logically equivalent to ~p --> q


The last column was supposed to be p ∨ q

T
T
T
F
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ghostwalker
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(Original post by khanpatel321)
I'm asking this

is p ∨ q logically equivalent to ~p --> q

So why are you doing a table for p \wedge q? Rather than p \vee q ? The first is "and" which is what you did in your table, and the second is "or".

Edit: And some of the entries in your ~p-->q column are incorrrect. It should run T,T,T,F.
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davros
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(Original post by khanpatel321)
I'm asking this

is p q logically equivalent to ~p --> q

Those are 2 different things!

If you're asking about p v q, why does your truth table have p AND q at the end???
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khanpatel321
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(Original post by ghostwalker)
So why are you doing a table for p \wedge q? Rather than p \vee q ? The first is "and" which is what you did in your table, and the second is "or".
(Original post by davros)
Those are 2 different things!

If you're asking about p v q, why does your truth table have p AND q at the end???
Check the edit
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davros
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(Original post by khanpatel321)
I'm asking this

is p q logically equivalent to ~p --> q


The last column was supposed to be p ∨ q

T
T
T
F
This is getting silly now!

Your column for -p -> q is wrong - remember that "a false proposition implies any proposition" so when -p is F you should have T in the -p -> q column
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ghostwalker
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(Original post by khanpatel321)
Check the edit
See the edit on my last post.
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khanpatel321
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(Original post by ghostwalker)
See the edit on my last post.
(Original post by davros)
This is getting silly now!

Your column for -p -> q is wrong - remember that "a false proposition implies any proposition" so when -p is F you should have T in the -p -> q column
Yeah sorry sorry I did the table wrong all along

Thank you
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Doctor_Einstein
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(Original post by khanpatel321)
Yeah sorry sorry I did the table wrong all along

Thank you
As I said, learn to do these quickly in your head using pure reasoning, then you'll never go wrong because you know the answer before you even draw the truth table.
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