Metamath Proof Explorer


Theorem cbvoprab13vw

Description: Change the first and third bound variables in an operation abstraction, using implicit substitution. (Contributed by GG, 14-Aug-2025)

Ref Expression
Hypothesis cbvoprab13vw.1 ⊢ x = w ∧ z = v → ψ ↔ χ
Assertion cbvoprab13vw ⊢ x y z | ψ = w y v | χ

Proof

Step Hyp Ref Expression
1 cbvoprab13vw.1 ⊢ x = w ∧ z = v → ψ ↔ χ
2 opeq1 ⊢ x = w → x y = w y
3 2 adantr ⊢ x = w ∧ z = v → x y = w y
4 simpr ⊢ x = w ∧ z = v → z = v
5 3 4 opeq12d ⊢ x = w ∧ z = v → x y z = w y v
6 5 eqeq2d ⊢ x = w ∧ z = v → t = x y z ↔ t = w y v
7 6 1 anbi12d ⊢ x = w ∧ z = v → t = x y z ∧ ψ ↔ t = w y v ∧ χ
8 7 cbvexdvaw ⊢ x = w → ∃ z t = x y z ∧ ψ ↔ ∃ v t = w y v ∧ χ
9 8 exbidv ⊢ x = w → ∃ y ∃ z t = x y z ∧ ψ ↔ ∃ y ∃ v t = w y v ∧ χ
10 9 cbvexvw ⊢ ∃ x ∃ y ∃ z t = x y z ∧ ψ ↔ ∃ w ∃ y ∃ v t = w y v ∧ χ
11 10 abbii ⊢ t | ∃ x ∃ y ∃ z t = x y z ∧ ψ = t | ∃ w ∃ y ∃ v t = w y v ∧ χ
12 df-oprab ⊢ x y z | ψ = t | ∃ x ∃ y ∃ z t = x y z ∧ ψ
13 df-oprab ⊢ w y v | χ = t | ∃ w ∃ y ∃ v t = w y v ∧ χ
14 11 12 13 3eqtr4i ⊢ x y z | ψ = w y v | χ