Chapter 2: Problem 27
The first stage of the method described to "push negations inward" was a method to climinate \(\rightarrow\) 's and \(\leftrightarrow\) 's. Prove that in the method to eliminate them, the process of substituting always stops, Consider, for example, the substitution in the formula $$ (p \leftrightarrow q) \leftrightarrow(r \leftrightarrow s) $$ If the substitution is first performed on the second \(\leftrightarrow\), the resultant formula is $$ ((p \leftrightarrow q) \rightarrow(r \leftrightarrow s)) \wedge((r \leftrightarrow s) \rightarrow(p \leftrightarrow q)) $$ which has more \(\leftrightarrow\) 's to replace than in the original formula! At first sight, one might expect that if the substitutions are made in the wrong order, the process might continue generating more \(\leftrightarrow\) 's at each stage, and the process might continue forever. (Hint: One method is to, instead of just counting the number of \(\leftrightarrow\) symbols, put a weight on each \(\leftrightarrow\) symbol, with the weight of the \(\leftrightarrow\) symbol in \(\psi \leftrightarrow x\) being dependent on the number of \(\leftrightarrow\) 's in \(\psi\) and \(\chi\). If the correct method of calculating weights is used, it can be shown that the total weight of the \(\leftrightarrow\) 's decreases with each substitution.
Short Answer
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