Metamath Proof Explorer

Theorem cbvraldva2

Description: Rule used to change the bound variable in a restricted universal quantifier with implicit substitution which also changes the quantifier domain. Deduction form. (Contributed by David Moews, 1-May-2017)

		Ref	Expression
	Hypotheses	cbvraldva2.1	$⊢ (φ \land x = y) \to (ψ \leftrightarrow χ)$
		cbvraldva2.2	$⊢ (φ \land x = y) \to A = B$
	Assertion	cbvraldva2	$⊢ φ \to (\forall x \in A ψ \leftrightarrow \forall y \in B χ)$

Proof

Step	Hyp	Ref	Expression
1		cbvraldva2.1	$⊢ (φ \land x = y) \to (ψ \leftrightarrow χ)$
2		cbvraldva2.2	$⊢ (φ \land x = y) \to A = B$
3		simpr	$⊢ (φ \land x = y) \to x = y$
4	3 2	eleq12d	$⊢ (φ \land x = y) \to (x \in A \leftrightarrow y \in B)$
5	4 1	imbi12d	$⊢ (φ \land x = y) \to ((x \in A \to ψ) \leftrightarrow (y \in B \to χ))$
6	5	expcom	$⊢ x = y \to (φ \to ((x \in A \to ψ) \leftrightarrow (y \in B \to χ)))$
7	6	pm5.74d	$⊢ x = y \to ((φ \to (x \in A \to ψ)) \leftrightarrow (φ \to (y \in B \to χ)))$
8	7	cbvalvw	$⊢ \forall x (φ \to (x \in A \to ψ)) \leftrightarrow \forall y (φ \to (y \in B \to χ))$
9		19.21v	$⊢ \forall x (φ \to (x \in A \to ψ)) \leftrightarrow (φ \to \forall x (x \in A \to ψ))$
10		19.21v	$⊢ \forall y (φ \to (y \in B \to χ)) \leftrightarrow (φ \to \forall y (y \in B \to χ))$
11	8 9 10	3bitr3i	$⊢ (φ \to \forall x (x \in A \to ψ)) \leftrightarrow (φ \to \forall y (y \in B \to χ))$
12	11	pm5.74ri	$⊢ φ \to (\forall x (x \in A \to ψ) \leftrightarrow \forall y (y \in B \to χ))$
13		df-ral	$⊢ \forall x \in A ψ \leftrightarrow \forall x (x \in A \to ψ)$
14		df-ral	$⊢ \forall y \in B χ \leftrightarrow \forall y (y \in B \to χ)$
15	12 13 14	3bitr4g	$⊢ φ \to (\forall x \in A ψ \leftrightarrow \forall y \in B χ)$