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Currently, I am self-teaching myself formal logic from E.J. Lemmon's Beginning Logic. So far, the book is discussing the rules of derivation for propositional logic, and I've struggled to fully understand the conditional proof rule.

The book gives a proof of P->Q |- (-Q)->(-P)

1 (1) P->Q assumption

2 (2) -Q assumption

1,2 (3) -P 1, 2 modus tollens

1 (4) -Q-> -P 2, 3 Conditional Proof

Since the negation of P is obtained by assuming the negation of Q, the negation of Q implies the negation of P; I understand how the rule allows one to put an implication into a conditional. However, what I fail to understand is why the assumption of 2 is discharged. Isn't it necessary to assume the negation of Q? Why does line 4 not depend upon the assumption of -Q? Lemmons says that when one uses conditional proof, there is always a discharged assumption. Why can one discharge an assumption in proving a conditional?

  • This is much less confusing proved by contradiction, incidentally. X -> Y can be false only if X is false and Y is true. If ~Q is false, Q is true. If ~P is true, P is false. So if ~Q -> ~P is false, then P -> Q is false -> true, which is false. Contradiction with assumption P -> Q. Therefore, ~Q -> ~P. – Rex Kerr Feb 2 '15 at 0:22
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There's two ways of getting out of sub-arguments.

One is proving that their outcomes are contradictory. This proves the opposite in the frame of the assumptions of the main argument.

The other is to discharge it as a conditional. In other words, you keep that this is an assumption but you transpose it as an instance of conditional. (This seems way harder reading the words than it is).

Or to put it another way, there's no difference between making a subargument and seeing what would be the case were the assumption of your subargument true. The conditional is in a sense just a short hand for a sub argument -- which under modus ponens of that assumption would obtain.

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