2.5.13: Soundness with Identity predicate
Natural deduction with rules for \(\eq[][]\) is sound.
Proof. Any formula of the form \(\eq[t][t]\) is valid, since for every structure \(\Struct M\) , \(\Sat{M}{\eq[t][t]}\) . (Note that we assume the term \(t\) to be ground, i.e., it contains no variables, so variable assignments are irrelevant).
Suppose the last inference in a derivation is \(\Elim{\eq[][]}\) , i.e., the derivation has the following form:
The premises \(\eq[t_1][t_2]\) and \(A(t_1)\) are derived from undischarged assumptions \(\Gamma_1\) and \(\Gamma_2\) , respectively. We want to show that \(A(t_2)\) follows from \(\Gamma_1 \cup \Gamma_2\) . Consider a structure \(\Struct{M}\) with \(\Sat{M}{\Gamma_1 \cup \Gamma_2}\) . By induction hypothesis, \(\Sat{M}{A(t_1)}\) and \(\Sat{M}{\eq[t_1][t_2]}\) . Therefore, \(\Value{t_1}{M} = \Value{t_2}{M}\) . Let \(s\) be any variable assignment, and \(s'\) be the \(x\) -variant given by \(s'(x) = \Value{t_1}{M} = \Value{t_2}{M}\) . By Proposition 5.13.3 , \(\Sat[,s]{M}{A(t_1)}\) iff \(\Sat[,s']{M}{A(x)}\) iff \(\Sat[,s]{M}{A(t_2)}\) . Since \(\Sat{M}{A(t_1)}\) , we have \(\Sat{M}{A(t_2)}\) . ◻