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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 → Ⅎ 𝑥 𝐴 )
	nfraldw.3	⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 )
Assertion	nfrexdw	⊢ ( 𝜑 → Ⅎ 𝑥 ∃ 𝑦 ∈ 𝐴 𝜓 )

Proof

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