Metamath Proof Explorer

Theorem cbvoprab23davw

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

		Ref	Expression
	Hypothesis	cbvoprab23davw.1	$⊢ ((φ \land y = w) \land z = v) \to (ψ \leftrightarrow χ)$
	Assertion	cbvoprab23davw	$⊢ φ \to \{⟨⟨x, y⟩, z⟩ \| ψ\} = \{⟨⟨x, w⟩, v⟩ \| χ\}$

Proof

Step	Hyp	Ref	Expression
1		cbvoprab23davw.1	$⊢ ((φ \land y = w) \land z = v) \to (ψ \leftrightarrow χ)$
2		eqidd	$⊢ ((φ \land y = w) \land z = v) \to x = x$
3		simplr	$⊢ ((φ \land y = w) \land z = v) \to y = w$
4	2 3	opeq12d	$⊢ ((φ \land y = w) \land z = v) \to ⟨x, y⟩ = ⟨x, w⟩$
5		simpr	$⊢ ((φ \land y = w) \land z = v) \to z = v$
6	4 5	opeq12d	$⊢ ((φ \land y = w) \land z = v) \to ⟨⟨x, y⟩, z⟩ = ⟨⟨x, w⟩, v⟩$
7	6	eqeq2d	$⊢ ((φ \land y = w) \land z = v) \to (t = ⟨⟨x, y⟩, z⟩ \leftrightarrow t = ⟨⟨x, w⟩, v⟩)$
8	7 1	anbi12d	$⊢ ((φ \land y = w) \land z = v) \to ((t = ⟨⟨x, y⟩, z⟩ \land ψ) \leftrightarrow (t = ⟨⟨x, w⟩, v⟩ \land χ))$
9	8	cbvexdvaw	$⊢ (φ \land y = w) \to (\exists z (t = ⟨⟨x, y⟩, z⟩ \land ψ) \leftrightarrow \exists v (t = ⟨⟨x, w⟩, v⟩ \land χ))$
10	9	cbvexdvaw	$⊢ φ \to (\exists y \exists z (t = ⟨⟨x, y⟩, z⟩ \land ψ) \leftrightarrow \exists w \exists v (t = ⟨⟨x, w⟩, v⟩ \land χ))$
11	10	exbidv	$⊢ φ \to (\exists x \exists y \exists z (t = ⟨⟨x, y⟩, z⟩ \land ψ) \leftrightarrow \exists x \exists w \exists v (t = ⟨⟨x, w⟩, v⟩ \land χ))$
12	11	abbidv	$⊢ φ \to \{t \| \exists x \exists y \exists z (t = ⟨⟨x, y⟩, z⟩ \land ψ)\} = \{t \| \exists x \exists w \exists v (t = ⟨⟨x, w⟩, v⟩ \land χ)\}$
13		df-oprab	$⊢ \{⟨⟨x, y⟩, z⟩ \| ψ\} = \{t \| \exists x \exists y \exists z (t = ⟨⟨x, y⟩, z⟩ \land ψ)\}$
14		df-oprab	$⊢ \{⟨⟨x, w⟩, v⟩ \| χ\} = \{t \| \exists x \exists w \exists v (t = ⟨⟨x, w⟩, v⟩ \land χ)\}$
15	12 13 14	3eqtr4g	$⊢ φ \to \{⟨⟨x, y⟩, z⟩ \| ψ\} = \{⟨⟨x, w⟩, v⟩ \| χ\}$