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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 → Ⅎ 𝑥 𝐴 )
	nfrald.3	⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 )
Assertion	nfrexd	⊢ ( 𝜑 → Ⅎ 𝑥 ∃ 𝑦 ∈ 𝐴 𝜓 )

Proof

Step	Hyp	Ref	Expression
1		nfrald.1	⊢ Ⅎ 𝑦 𝜑
2		nfrald.2	⊢ ( 𝜑 → Ⅎ 𝑥 𝐴 )
3		nfrald.3	⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 )
4		dfrex2	⊢ ( ∃ 𝑦 ∈ 𝐴 𝜓 ↔ ¬ ∀ 𝑦 ∈ 𝐴 ¬ 𝜓 )
5	3	nfnd	⊢ ( 𝜑 → Ⅎ 𝑥 ¬ 𝜓 )
6	1 2 5	nfrald	⊢ ( 𝜑 → Ⅎ 𝑥 ∀ 𝑦 ∈ 𝐴 ¬ 𝜓 )
7	6	nfnd	⊢ ( 𝜑 → Ⅎ 𝑥 ¬ ∀ 𝑦 ∈ 𝐴 ¬ 𝜓 )
8	4 7	nfxfrd	⊢ ( 𝜑 → Ⅎ 𝑥 ∃ 𝑦 ∈ 𝐴 𝜓 )