Theorem spcegf 2936

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 1793	. . . 4 ⊢ Ⅎx ¬ ψ
4		spcgf.3	. . . . 5 ⊢ (x = A → (φ ↔ ψ))
5	4	notbid 285	. . . 4 ⊢ (x = A → (¬ φ ↔ ¬ ψ))
6	1, 3, 5	spcgf 2935	. . 3 ⊢ (A ∈ V → (∀x ¬ φ → ¬ ψ))
7	6	con2d 107	. 2 ⊢ (A ∈ V → (ψ → ¬ ∀x ¬ φ))
8		df-ex 1542	. 2 ⊢ (∃xφ ↔ ¬ ∀x ¬ φ)
9	7, 8	syl6ibr 218	1 ⊢ (A ∈ V → (ψ → ∃xφ))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176  ∀wal 1540  ∃wex 1541  Ⅎwnf 1544   = wceq 1642   ∈ wcel 1710  Ⅎwnfc 2477

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862

This theorem is used by:  spcegv  2941  rspce  2951