2.5.12: Derivations with Identity predicate

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Derivations with identity predicate require additional inference rules.

In the above rules, \(t\), \(t_1\), and \(t_2\) are closed terms. The \(\Intro{\eq[][]}\) rule allows us to derive any identity statement of the form \(\eq[t][t]\) outright, from no assumptions.

Example \(\PageIndex{1}\)

If \(s\) and \(t\) are closed terms, then \(A(s), \eq[s][t] \Proves A(t)\):

This may be familiar as the “principle of substitutability of identicals,” or Leibniz’ Law.

Problem \(\PageIndex{1}\)

Prove that \(=\) is both symmetric and transitive, i.e., give derivations of \(\lforall{x}{\lforall{y}{(\eq[x][y] \lif \eq[y][x])}}\) and \(\lforall{x}{\lforall{y}{\lforall{z}{}((\eq[x][y] \land \eq[y][z]) \lif \eq[x][z])}}\)

Example \(\PageIndex{2}\)

We derive the sentence

\[\lforall{x}{\lforall{y}{((A(x) \land A(y)) \lif \eq[x][y])}} \nonumber\]

from the sentence

\[\lexists{x}{\lforall{y}{(A(y) \lif \eq[y][x])}}\nonumber\]

We develop the derivation backwards:

We’ll now have to use the main assumption: since it is an existential formula, we use \(\Elim{\lexists{}{}}\) to derive the intermediary conclusion \(\eq[a][b]\).

The sub-derivation on the top right is completed by using its assumptions to show that \(\eq[a][c]\) and \(\eq[b][c]\). This requires two separate derivations. The derivation for \(\eq[a][c]\) is as follows:

From \(\eq[a][c]\) and \(\eq[b][c]\) we derive \(\eq[a][b]\) by \(\Elim{\eq[][]}\).

Problem \(\PageIndex{2}\)

Give derivations of the following formulas:

1. \(\lforall{x}{\lforall{y}{((\eq[x][y] \land A(x)) \lif A(y))}}\)
2. \(\lexists{x}{A(x)} \land \lforall{y}{\lforall{z}{((A(y) \land A(z)) \lif \eq[y][z])}} \lif \lexists{x}{(A(x) \land \lforall{y}{(A(y) \lif \eq[y][x])})}\)