Two practice exercises in Natural Deductive Logic: RULES #3 (=I, =E) - Logic

We do two practice exercises on natural deductive proofs using biconditional elimination, biconditional introduction, and all of our rules from before. These proofs are quite long, between 15 and 25 lines each!

#PropositionalLogic #LogicProofs

0:00 [Intro]
0:20 [Question #1]
9:02 [Question #2]

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In this video on #Logic, we learn our last two rules for proofs. We learn biconditional introduction and biconditional elimination. Then we do a few example proofs.

[khaled]

I think the first example has a contradictory, because we are trying to get to (q and not p ) but on the line 8 there's p<>q, which means if q is true p must be true so then not p is false, so (q and not p ) will never be true...
Also line 3 we have not S but on line 11 we found S.

[yannsalmon]

So cool ! I love the fact that you put yourself in danger.
The first video I really block, so that's nice to see that I'm not alone :)

[chipperMDW]

Suggested modifications to the first proof:

Cross out line 16; on line 17, you can reiterate S from line 11 instead of proving it a second time with MP.

But then there was no need to assume ~Q in particular on line 15. You could have assumed anything you wanted and still reiterated S and ~S. So, instead, just assume ~(Q & ~P) on line 15 and replace Q in lines 19 and 20 accordingly. (Note that, while you're in the subproof that assumes R, you can always derive that contradiction, so you can conclude anything via that method of RAA and DN.)

At that point, you'll already have Q & ~P on line 20, so you can cross out line 21. Also, there'll be no need to show ~P on its own anymore, so you can cross out lines 12-14.

[noobygames]

Imagine hypothesis is R = true, now P <-> Q is true. this means that P-> Q and Q->P right? So isnt it impossible to have Q ^ ~P here ? because if Q is true then P is also true so it contradicts... I dont see any other comments about this so I guess im wrong, but can someone explain me where I go wrong here?

[janezperme]

Regarding the second example: I don't understand why can one assume p, yet there is a hypothesis ~p & ~q, which means ~p, after conjuntion elimination. So p and ~p ? Wouldn't that be an immediate contradiction ? Many thanks for your thoughts.

[techboyzeeshan]

can someone please clarify why he used contradiction to get Q coz from line 3 and 11 we already have a contradiction that our assumption of R is wrong so that means we should not proceed with the remaining proof as we can't assume R to prove the conditional given in the conclusion

[yannsalmon]

I wondered: the fact that we can add new hypotheses seems weird to me because the end result is not really proved by the initial hypotheses, but rather by the initial and the new ones.
Am I missing something ?