Metamath Proof Explorer


Theorem sbiedw

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbiev ). Version of sbied with a disjoint variable condition, requiring fewer axioms. (Contributed by NM, 30-Jun-1994) Avoid ax-13 . (Revised by GG, 10-Jan-2024)

Ref Expression
Hypotheses sbiedw.1 ⊢ Ⅎ x φ
sbiedw.2 ⊢ φ → Ⅎ x χ
sbiedw.3 ⊢ φ → x = y → ψ ↔ χ
Assertion sbiedw ⊢ φ → y x ψ ↔ χ

Proof

Step Hyp Ref Expression
1 sbiedw.1 ⊢ Ⅎ x φ
2 sbiedw.2 ⊢ φ → Ⅎ x χ
3 sbiedw.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 sbiev ⊢ 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