Theorem spcegf 3550

Description: Existential specialization, using implicit substitution. (Contributed by NM, 2-Feb-1997.)

Hypotheses

Ref	Expression
spcgf.1	⊢ Ⅎ𝑥𝐴
spcgf.2	⊢ Ⅎ𝑥𝜓
spcgf.3	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
spcegf	⊢ (𝐴 ∈ 𝑉 → (𝜓 → ∃𝑥𝜑))

Proof of Theorem spcegf

Step	Hyp	Ref	Expression
1		spcgf.1	. . . 4 ⊢ Ⅎ𝑥𝐴
2		spcgf.2	. . . . 5 ⊢ Ⅎ𝑥𝜓
3	2	nfn 1889	. . . 4 ⊢ Ⅎ𝑥 ¬ 𝜓
4		spcgf.3	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
5	4	notbid 321	. . . 4 ⊢ (𝑥 = 𝐴 → (¬ 𝜑 ↔ ¬ 𝜓))
6	1, 3, 5	spcgf 3549	. . 3 ⊢ (𝐴 ∈ 𝑉 → (∀𝑥 ¬ 𝜑 → ¬ 𝜓))
7	6	con2d 135	. 2 ⊢ (𝐴 ∈ 𝑉 → (𝜓 → ¬ ∀𝑥 ¬ 𝜑))
8		df-ex 1812	. 2 ⊢ (∃𝑥𝜑 ↔ ¬ ∀𝑥 ¬ 𝜑)
9	7, 8	imbitrrdi 255	1 ⊢ (𝐴 ∈ 𝑉 → (𝜓 → ∃𝑥𝜑))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209  ∀wal 1567   = wceq 1569  ∃wex 1811  Ⅎwnf 1815   ∈ wcel 2146  Ⅎwnfc 2911

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2196  ax-12 2215  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1812  df-nf 1816  df-cleq 2755  df-clel 2839  df-nfc 2913

This theorem is used by:  rspce  3569  euotd  5490  bnj607  35602  bnj1491  35743  permaxrep  46138  rspcegf  46182  stoweidlem36  47188  stoweidlem46  47198  ichnreuop  48696  ichreuopeq  48697