[Math] Determining if Argument is Valid via Short-Cut Method

discrete mathematicslogic

I understand there are two ways to determine validity of an argument. The first way is to construct a truth table and if the statement consisting of the premises combined together implying the conclusion is a tautology, then it is valid.

The other method is to use the fact that it no valid arguments have all premises to be true and the conclusion to be false. So we could assume that the conclusion is false, and work are way through the premises, determining their truth. This is the "short cut" method.

With regards to this question, how could you use the short cut method?

P

Q -> (R->S)

U -> R

P V Q


~S -> ~U

So we would assume ~S -> ~U is false. Does that mean that S -> U would be true? Where would we take it from here.

Best Answer

You last comment is right: the argument is invalid.

An argument is valid if and only if there is no possible situation in which the premisses would be true and the conclusion false.

From this, it follows that

an argument with true premisses and a false conclusion can't be valid.

What you call "the short cut method" is a semantical proof by contradiction, i.e. assuming the falsity of the conclusion and trying to derive a contradiction.


Thus, we have to assume that $\lnot S \to \lnot U$ is false (which does not means that $S \to U$ is true...), i.e.:

$\lnot S$ true and $\lnot U$ false, i.e. $S$ false and $U$ true.

The premises must be all true; thus $P$ (first premise) is true and this is enough to satisfy also the fourth one: $P \lor Q$.

Assuming $U$ true implies that, with the third premise: $U \to R$, that also $R$ must be true.

Consider now the second premise: $Q \to (R \to S)$.

Up to now we have:

$S$ false, $U$ true, $P$ true and $R$ true.

Thus: $R \to S$ is false and this implies $Q$ false.

This is "compatible" with the satisfaction of $P \lor Q$, and thus we have found no contradiction, i.e. the argument is invalid.

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