Theorem spcgf 3545

Description: Rule of specialization, using implicit substitution. Compare Theorem 7.3 of [Quine] p. 44. (Contributed by NM, 2-Feb-1997.) (Revised by Andrew Salmon, 12-Aug-2011.)

Hypotheses

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

Assertion

Ref	Expression
spcgf	⊢ (𝐴 ∈ 𝑉 → (∀𝑥𝜑 → 𝜓))

Proof of Theorem spcgf

Step	Hyp	Ref	Expression
1		spcgf.2	. . 3 ⊢ Ⅎ𝑥𝜓
2		spcgf.1	. . 3 ⊢ Ⅎ𝑥𝐴
3	1, 2	spcgft 3512	. 2 ⊢ (∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) → (𝐴 ∈ 𝑉 → (∀𝑥𝜑 → 𝜓)))
4		spcgf.3	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
5	3, 4	mpg 1830	1 ⊢ (𝐴 ∈ 𝑉 → (∀𝑥𝜑 → 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568   = wceq 1570  Ⅎwnf 1816   ∈ wcel 2145  Ⅎwnfc 2907

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817  df-cleq 2752  df-clel 2835  df-nfc 2909

This theorem is used by:  spcegf  3546  rspc  3564  eusvnf  5353  gropd  29542  grstructd  29543