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

[Phatfoo]

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.

 

[Dr.Peterson]

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]

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!