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I am (re-)working through the Gensler logic book to refresh my command of formal logic. For the most part, he is using proof by contradiction to achieve results. I noticed that the proofs I am writing are somewhat different from his own. Typically, for a set of premises S and a conclusion Q, he assumes ~Q and then from there derives both A and ~A for some A, thereby proving Q. I realized that I have been assuming ~Q just as he has, but instead of finding a complementary A and ~A from there to make my contradiction, I have instead proved that Q is true from S and ~Q. Is this also contradictory, or am I missing something?

orangeskid
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readyready15728
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That's fine: you're proving $\neg Q\implies Q$, which implies $\neg Q\implies Q\wedge \neg Q$, so that $\neg Q$ again implies a contradiction.

Kevin Carlson
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  • HALLELUJAH! My question then is why no one else seems to draw attention to this in the logic texts that I've found. – readyready15728 Sep 17 '14 at 23:50
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    Eh, I suppose just because it's a special case of the proof by contradiction they already cover. I'll comment that quite frequently people write proofs that start out by assuming $\neg Q$, prove $Q$ without every using the assumption $\neg Q$, and then conclude $Q$ "by contradiction!" So I'd have a look at your proofs to be sure the apparent contradiction you're deriving isn't really just disguising a direct proof. – Kevin Carlson Sep 17 '14 at 23:53
  • I'm definitely not writing direct proofs here. I remember when I read Copi some time ago, but then forgot everything in there, he taught direct proofs. Gensler is all about indirect proofs and all the proofs I'm writing ultimately rely on the assumption of the negation of the thing to be prove that I'm making. I was sweating bullets for a moment because, if I had been mistaken, all my work would have been undone. – readyready15728 Sep 17 '14 at 23:57
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I would like to add some comments on this question from a formal point of view.

In mathematical logic, one way to formalize the law of contradiction is through the propositional law :

$(A → B) → ((A → ¬B) → ¬A)$ --- (*).

This form is intuitionistically valid; it is enough to change it a little bit, to have classical logic :

$(\lnot A → B) → ((\lnot A → \lnot B) → A)$ --- (§).

This law is easily derivable from (*) through Double Negation : $\lnot \lnot A \rightarrow A$, which is not intuitionistically valid.

Having said that, the law we are discussing :

$(\lnot Q \rightarrow Q) \rightarrow Q$

is itself easily derivable from the law of contradiction :

i) $\lnot Q \rightarrow Q$ --- assumed

ii) $(\lnot Q → Q) → ((\lnot Q → \lnot Q) → Q)$ --- from (§) with $Q$ in place of both $A,B$ [this step amounts to applying Henning's comment above : "just taking $A$ to be $Q$"]

iii) $(\lnot Q → \lnot Q) → Q$ --- from i) and ii) by modus ponens

iv) $Q$ --- from iii) and the logic law : $A \rightarrow A$, by modus ponens

v) $(\lnot Q \rightarrow Q) \rightarrow Q$ --- from i) and iv) by Deduction Theorem.

Thus, we are licensed to say that it is also a form of the law of contradiction.

Mauro ALLEGRANZA
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