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Metamath Proof Explorer

Theorem nfrexdw

Description: Deduction version of nfrexw . (Contributed by Mario Carneiro, 14-Oct-2016) Add disjoint variable condition to avoid ax-13 . See nfrexd for a less restrictive version requiring more axioms. (Revised by GG, 20-Jan-2024)

	Ref	Expression
Hypotheses	nfraldw.1	$⊢ Ⅎ y φ$
	nfraldw.2	$⊢ φ \to \underline{Ⅎ} x A$
	nfraldw.3	$⊢ φ \to Ⅎ x ψ$
Assertion	nfrexdw	$⊢ φ \to Ⅎ x \exists y \in A ψ$

Proof

Step	Hyp	Ref	Expression
1		nfraldw.1	$⊢ Ⅎ y φ$
2		nfraldw.2	$⊢ φ \to \underline{Ⅎ} x A$
3		nfraldw.3	$⊢ φ \to Ⅎ x ψ$
4		dfrex2	$⊢ \exists y \in A ψ \leftrightarrow \neg \forall y \in A \neg ψ$
5	3	nfnd	$⊢ φ \to Ⅎ x \neg ψ$
6	1 2 5	nfraldw	$⊢ φ \to Ⅎ x \forall y \in A \neg ψ$
7	6	nfnd	$⊢ φ \to Ⅎ x \neg \forall y \in A \neg ψ$
8	4 7	nfxfrd	$⊢ φ \to Ⅎ x \exists y \in A ψ$