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Metamath Proof Explorer

Theorem nfrexd

Description: Deduction version of nfrex . Usage of this theorem is discouraged because it depends on ax-13 . See nfrexdw for a version with a disjoint variable condition, but not requiring ax-13 . (Contributed by Mario Carneiro, 14-Oct-2016) (New usage is discouraged.)

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

Proof

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