Metamath Proof Explorer


Theorem sbequbidv

Description: Deduction substituting both sides of a biconditional. (Contributed by GG, 1-Sep-2025)

Ref Expression
Hypotheses sbequbidv.1 ⊢ φ → u = v
sbequbidv.2 ⊢ φ → ψ ↔ χ
Assertion sbequbidv ⊢ φ → u x ψ ↔ v x χ

Proof

Step Hyp Ref Expression
1 sbequbidv.1 ⊢ φ → u = v
2 sbequbidv.2 ⊢ φ → ψ ↔ χ
3 equequ2 ⊢ u = v → t = u ↔ t = v
4 1 3 syl ⊢ φ → t = u ↔ t = v
5 2 imbi2d ⊢ φ → x = t → ψ ↔ x = t → χ
6 5 albidv ⊢ φ → ∀ x x = t → ψ ↔ ∀ x x = t → χ
7 4 6 imbi12d ⊢ φ → t = u → ∀ x x = t → ψ ↔ t = v → ∀ x x = t → χ
8 7 albidv ⊢ φ → ∀ t t = u → ∀ x x = t → ψ ↔ ∀ t t = v → ∀ x x = t → χ
9 dfsb ⊢ u x ψ ↔ ∀ t t = u → ∀ x x = t → ψ
10 dfsb ⊢ v x χ ↔ ∀ t t = v → ∀ x x = t → χ
11 8 9 10 3bitr4g ⊢ φ → u x ψ ↔ v x χ