# Section 1.7: Problem 6 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
Let $<\alpha_{0},\ldots,\alpha_{n}>$ be a deduction from a set $\Sigma$ of wffs, as in the preceding problem. Show that $\Sigma\vDash\alpha_{k}$ for each $k\le n$ . Suggestion: Use strong induction on k, so that the inductive hypothesis is that $\Sigma\vDash\alpha_{i}$ for all $i .
Let $D_{k}=\{\alpha_{i}|i , i.e. the set of the first $k$ wffs forming the deduction. We show that if $\Sigma\vDash D_{k}$ then $\Sigma\vDash\alpha_{k}$ for each $k\le n$ , and, in particular, $\Sigma\vDash D_{k+1}=D_{k};\alpha_{k}$ . Since $D_{0}=\emptyset$ , $\Sigma\vDash D_{0}$ . Suppose that $\Sigma\vDash D_{k}$ . Then if $\alpha_{k}$ is a tautology, then $\Sigma\vDash\alpha_{k}$ , if $\alpha_{k}\in\Sigma$ , then $\Sigma\vDash\alpha_{k}$ , and, finally, if is such that $\alpha_{i}=(\alpha_{j}\rightarrow\alpha_{k})$ for some $i,j , then $\Sigma\vDash(\alpha_{j}\rightarrow\alpha_{k})$ implying (Exercise 4(a), Section 1.2) $\Sigma;\alpha_{j}=\Sigma\vDash\alpha_{k}$ .