Metamath Proof Explorer

Theorem sbied

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbie ) Usage of this theorem is discouraged because it depends on ax-13 . See sbiedw , sbiedvw for variants using disjoint variables, but requiring fewer axioms. (Contributed by NM, 30-Jun-1994) (Revised by Mario Carneiro, 4-Oct-2016) (Proof shortened by Wolf Lammen, 24-Jun-2018) (New usage is discouraged.)

Ref Expression
Hypotheses sbied.1 x φ
sbied.2 φ x χ
sbied.3 φ x = y ψ χ
Assertion sbied φ y x ψ χ


Step Hyp Ref Expression
1 sbied.1 x φ
2 sbied.2 φ x χ
3 sbied.3 φ x = y ψ χ
4 1 sbrim y x φ ψ φ y x ψ
5 1 2 nfim1 x φ χ
6 3 com12 x = y φ ψ χ
7 6 pm5.74d x = y φ ψ φ χ
8 5 7 sbie y x φ ψ φ χ
9 4 8 bitr3i φ y x ψ φ χ
10 9 pm5.74ri φ y x ψ χ