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Metamath Proof Explorer

Theorem cbvabv

Description: Rule used to change bound variables, using implicit substitution. Version of cbvab with disjoint variable conditions requiring fewer axioms. (Contributed by NM, 26-May-1999) Require x , y be disjoint to avoid ax-11 and ax-13 . (Revised by Steven Nguyen, 4-Dec-2022)

		Ref	Expression
	Hypothesis	cbvabv.1	$⊢ x = y \to (φ \leftrightarrow ψ)$
	Assertion	cbvabv	$⊢ \{x \| φ\} = \{y \| ψ\}$

Proof

Step	Hyp	Ref	Expression
1		cbvabv.1	$⊢ x = y \to (φ \leftrightarrow ψ)$
2	1	cbvsbv	$⊢ [z/ x] φ \leftrightarrow [z/ y] ψ$
3		df-clab	$⊢ z \in \{x \| φ\} \leftrightarrow [z/ x] φ$
4		df-clab	$⊢ z \in \{y \| ψ\} \leftrightarrow [z/ y] ψ$
5	2 3 4	3bitr4i	$⊢ z \in \{x \| φ\} \leftrightarrow z \in \{y \| ψ\}$
6	5	eqriv	$⊢ \{x \| φ\} = \{y \| ψ\}$