# Prove that a proposition which is based on a valid deduction is a tautology

Given a valid deduction rule, based on a set of propositions, e.g.:

$P_1$

$P_2$

$\vdots$

$P_n$

∴ $Q$

Prove that the statement:-

$(P_1 \land P_2 \land \ldots \land P_n) \to Q$.

is a tautology.

Now I know that when interpreting a deduction rule all premises in a deduction rule are conjoined together to imply the result. However this creates just a restatement of the latter statement which I'm trying to show is a tautology.

Is it really this simple? I.e. if I have two identical statements, and we know the first is true, then the second is true, so $T=T$.

Or do I have to prove the latter statement is a tautology using some knowledge from the deduction rule I haven't gathered?

• this is basically a restatement of the deduction theorem – Dan Jan 4 '18 at 11:15
• Apply the definition of valid deduction rule to show that the case $(P_1 \land \ldots \land P_n)$ TRUE and $Q$ FALSE leads to a contradiction. – Mauro ALLEGRANZA Jan 4 '18 at 11:17

Just use the definitions of what it means for an argument to be valid, and what it means for a statement to be a tautology, together with the truth-functional semantics of the connectives involved.

That is, by definition of a valid argument, the argument

$P_1$

$P_2$

$...$

$P_n$

$\therefore$

$Q$

is valid if and only if there is no truth-assignment that sets all of $P_1$ through $P_n$ to true and $Q$ to false.

By semantics of the $\land$, that means that there is no truth-assignment that sets $P_1 \land P_2 \land ... \land P_n$ to true and $Q$ to false.

By semantics of the $\rightarrow$, that means that there is no truth-assignment that sets $(P_1 \land P_2 \land ... \land P_n) \rightarrow Q$ to false.

And by definition of a statement to be a tautology, that means that $(P_1 \land P_2 \land ... \land P_n) \rightarrow Q$ is a tautology.