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Metamath Proof Explorer

Theorem equsexvw

Description: Version of equsexv with a disjoint variable condition, and of equsex with two disjoint variable conditions, which requires fewer axioms. See also the dual form equsalvw . (Contributed by BJ, 31-May-2019) (Proof shortened by Wolf Lammen, 23-Oct-2023)

		Ref	Expression
	Hypothesis	equsalvw.1	$⊢ x = y \to (φ \leftrightarrow ψ)$
	Assertion	equsexvw	$⊢ \exists x (x = y \land φ) \leftrightarrow ψ$

Proof

Step	Hyp	Ref	Expression
1		equsalvw.1	$⊢ x = y \to (φ \leftrightarrow ψ)$
2		alinexa	$⊢ \forall x (x = y \to \neg φ) \leftrightarrow \neg \exists x (x = y \land φ)$
3	1	notbid	$⊢ x = y \to (\neg φ \leftrightarrow \neg ψ)$
4	3	equsalvw	$⊢ \forall x (x = y \to \neg φ) \leftrightarrow \neg ψ$
5	2 4	bitr3i	$⊢ \neg \exists x (x = y \land φ) \leftrightarrow \neg ψ$
6	5	con4bii	$⊢ \exists x (x = y \land φ) \leftrightarrow ψ$