-1

An excerpt from Logic 2010:

enter image description here

In particular, what is confusing is that it permits assuming the conditional but then reaching a contradiction to prove the conditional. In my experience, that is not a proof by contradiction. Lines 7 - 10.

5
  • Is X valid? Commented Mar 26 at 2:14
  • can you explain what "id" means? Im assuming r is repeat, line 10 is a contradiction, line 11 is just a repeat of 3. not sure what line 12 is. But so far I dont see any conclusion, and how subproofs are handled (i.e. line 8-10) seems weird. Commented Mar 26 at 4:09
  • @ayylien id means indirect derivation. My main confusion is the subproof from 7 - 10, where we ass cd but end with an indirect derivation via a contradiction. Commented Mar 26 at 5:18
  • 1
    Quite confused... but in a nutshell you have to derive separately P and not-Q. Thus, two sub-proofs: (i) assume Q, derive P to Q and thus we have the first contra: conclude with not-Q discharging the assumption; (ii) assume not-P, derive P to Q and we have the second contra: derive not-not-P discharging the assumption and then conclude with Double Negation. Commented Mar 26 at 6:56
  • @user129393192 line 11 isn't an indirect derivation, it's just line 3 repeated (if r means repeat). The subproof 7-10 isn't actually being used after, and I doubt the proof is correct at all. Commented Mar 26 at 17:34

2 Answers 2

1

I don't really like the way that proof is presented, and I know this isn't what you asked, but if you want to prove ~(P -> Q) is mutually implicated with P ^ ~Q, we can just compare the truth tables.

P implies Q has a certain truth table

P Q P -> Q
T T T
T F F
F T T
F F T

Which means ~ (P -> Q) ought to have the opposite truth table.

P Q ~ (P -> Q)
T T F
T F T
F T F
F F F

The only row on that truth table is the row where P is true and Q is false.

0

That looks like something I used in an attempted proof recently on a question I asked here regarding moral relativism. It was not well received on this site by several other commenters.

I never used that software so I'm not familiar with it and can't really read what it's saying. I guess the aim is to show why NOT(if P then Q) is equivalent to (P and NOT Q). The first statement is implying that (if P then Q) is true, that means NOT(if P then Q) is false. (If P then Q) is false when P is assumed but it leads to something other than Q being true. That happens when P is true and Q is false, in other words, when (P and NOT Q) is true.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .