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 ⊢ Ⅎ x φ
sbied.2 ⊢ φ → Ⅎ x χ
sbied.3 ⊢ φ → x = y → ψ ↔ χ
Assertion sbied ⊢ φ → y x ψ ↔ χ

Proof

Step Hyp Ref Expression
1 sbied.1 ⊢ Ⅎ x φ
2 sbied.2 ⊢ φ → Ⅎ 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 ⊢ Ⅎ 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