A second truth table query: "[not-(not-(not-p))] -> q = p or q" ?

Phatfoo

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May 22, 2018
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16
As you can see, I've included my steps. The question is whether or not these statements are true or false.

ink (2).jpg

The second one got me a bit. I'm not entirely certain if "implies" is logically equivalent to "and".

If I've gone wrong, then please give me some advice. I'd love to get this correct.
 

Dr.Peterson

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Nov 12, 2017
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3,052
As you can see, I've included my steps. The question is whether or not these statements are true or false.

The second one got me a bit. I'm not entirely certain if "implies" is logically equivalent to "and".

If I've gone wrong, then please give me some advice. I'd love to get this correct.
No, "implies" is not equivalent to "and"; if it were, we wouldn't need two different symbols, would we?

Have you learned that \(\displaystyle p\rightarrow q \equiv \neg p \vee q \)?
 

Phatfoo

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Joined
May 22, 2018
Messages
16
No, "implies" is not equivalent to "and"; if it were, we wouldn't need two different symbols, would we?

Have you learned that \(\displaystyle p\rightarrow q \equiv \neg p \vee q \)?
Thank you for this. I did learn that but I forgot!
 
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