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The instructions state "using only results of Sec. 4 or earlier portions of the present exercise, prove:" We have shown already the following: (a) $\vdash P\vee Q \supset Q \vee P$ and (b) $\vdash P \supset P \vee Q$.

We still haven't establish associativity for the "or" connective, and we are supposed to assume association from left to right. So $P_1\vee P_2\vee \cdots \vee P_n$ is something like $(\cdots (((P_1 \vee P _2)\vee P_3)\vee \cdots)\vee P_n)$.

There are 29 Theorems that we have established in Sec. 4, I am hoping that the reader has access to the book. Two theorems that I think might be useful are:

Theorem IV.4.18. $\vdash P_1 P_2 \cdots P_n\supset P_m$ for $1\leq m\leq n$ , and Theorem IV.4.7. $\sim P \supset \sim Q \vdash Q \supset P$

But, I can't think of how to apply Theorem IV.4.18. to $P_1\vee P_2\vee \cdots \vee P_n$.

JimmyJackson
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1 Answers1

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Let's look at a special case: $n=4$, $m=3$. We want to show $\vdash P_3\supset (((P_1\vee P_2)\vee P_3)\vee P_4)$.

  • We know already that $\vdash P_3\supset ((P_1\vee P_2)\vee P_3)$.

  • And we know that $\vdash((P_1\vee P_2)\vee P_3)\supset(((P_1\vee P_2)\vee P_3)\supset P_4)$.

So all we need is "$\vdash A\supset B$ and $\vdash B\supset C$ implies $\vdash A\supset C$" . . .

The general problem is solved using essentially the same structure, plus induction.

Noah Schweber
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