Metamath Proof Explorer


Theorem cbvoprab2davw

Description: Change the second bound variable in an operation abstraction. Deduction form. (Contributed by GG, 14-Aug-2025)

Ref Expression
Hypothesis cbvoprab2davw.1 ⊢ φ ∧ y = w → ψ ↔ χ
Assertion cbvoprab2davw ⊢ φ → x y z | ψ = x w z | χ

Proof

Step Hyp Ref Expression
1 cbvoprab2davw.1 ⊢ φ ∧ y = w → ψ ↔ χ
2 opeq2 ⊢ y = w → x y = x w
3 2 adantl ⊢ φ ∧ y = w → x y = x w
4 3 opeq1d ⊢ φ ∧ y = w → x y z = x w z
5 4 eqeq2d ⊢ φ ∧ y = w → t = x y z ↔ t = x w z
6 5 1 anbi12d ⊢ φ ∧ y = w → t = x y z ∧ ψ ↔ t = x w z ∧ χ
7 6 exbidv ⊢ φ ∧ y = w → ∃ z t = x y z ∧ ψ ↔ ∃ z t = x w z ∧ χ
8 7 cbvexdvaw ⊢ φ → ∃ y ∃ z t = x y z ∧ ψ ↔ ∃ w ∃ z t = x w z ∧ χ
9 8 exbidv ⊢ φ → ∃ x ∃ y ∃ z t = x y z ∧ ψ ↔ ∃ x ∃ w ∃ z t = x w z ∧ χ
10 9 abbidv ⊢ φ → t | ∃ x ∃ y ∃ z t = x y z ∧ ψ = t | ∃ x ∃ w ∃ z t = x w z ∧ χ
11 df-oprab ⊢ x y z | ψ = t | ∃ x ∃ y ∃ z t = x y z ∧ ψ
12 df-oprab ⊢ x w z | χ = t | ∃ x ∃ w ∃ z t = x w z ∧ χ
13 10 11 12 3eqtr4g ⊢ φ → x y z | ψ = x w z | χ