cbvsbcw

Description: Change bound variables in a wff substitution. Version of cbvsbc with a disjoint variable condition, which does not require ax-13 . (Contributed by Jeff Hankins, 19-Sep-2009) Avoid ax-13 . (Revised by GG, 10-Jan-2024)

	Ref	Expression
Hypotheses	cbvsbcw.1	$⊢ Ⅎ y φ$
	cbvsbcw.2	$⊢ Ⅎ x ψ$
	cbvsbcw.3	$⊢ x = y \to (φ \leftrightarrow ψ)$
Assertion	cbvsbcw	$⊢ \underset{˙}{[} A / x \underset{˙}{]} φ \leftrightarrow \underset{˙}{[} A / y \underset{˙}{]} ψ$

Proof

Step	Hyp	Ref	Expression
1		cbvsbcw.1	$⊢ Ⅎ y φ$
2		cbvsbcw.2	$⊢ Ⅎ x ψ$
3		cbvsbcw.3	$⊢ x = y \to (φ \leftrightarrow ψ)$
4	1 2 3	cbvabw	$⊢ \{x \| φ\} = \{y \| ψ\}$
5	4	eleq2i	$⊢ A \in \{x \| φ\} \leftrightarrow A \in \{y \| ψ\}$
6		df-sbc	$⊢ \underset{˙}{[} A / x \underset{˙}{]} φ \leftrightarrow A \in \{x \| φ\}$
7		df-sbc	$⊢ \underset{˙}{[} A / y \underset{˙}{]} ψ \leftrightarrow A \in \{y \| ψ\}$
8	5 6 7	3bitr4i	$⊢ \underset{˙}{[} A / x \underset{˙}{]} φ \leftrightarrow \underset{˙}{[} A / y \underset{˙}{]} ψ$

Metamath Proof Explorer

Theorem cbvsbcw

Proof