Proof 01. 1 ⊢ x = y, hypo. 02. 1 ⊢ y = x, eq_subst 1 eq_refl, P u ↔ u = x. eq_symm. ⊢ x = y → y = x, subj_intro 2.
DependenciesThe given proof depends on two axioms:eq_refl, eq_subst.