Theorem spcv 2691

Description: Rule of specialization, using implicit substitution. (Contributed by NM, 22-Jun-1994.)

Hypotheses

Ref	Expression
spcv.1	⊢ 𝐴 ∈ V
spcv.2	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
spcv	⊢ (∀𝑥𝜑 → 𝜓)

Distinct variable groups:   𝑥,𝐴   𝜓,𝑥

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem spcv

Step	Hyp	Ref	Expression
1		spcv.1	. 2 ⊢ 𝐴 ∈ V
2		spcv.2	. . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
3	2	spcgv 2685	. 2 ⊢ (𝐴 ∈ V → (∀𝑥𝜑 → 𝜓))
4	1, 3	ax-mp 7	1 ⊢ (∀𝑥𝜑 → 𝜓)

Syntax hints:   → wi 4   ↔ wb 103  ∀wal 1282   = wceq 1284   ∈ wcel 1433  Vcvv 2601

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603

This theorem is referenced by:  morex  2776  rext  3970  ontr2exmid  4268  regexmidlem1  4276  reg2exmid  4279  relop  4504  rdgtfr  5984  ssfiexmid  6361  domfiexmid  6363  diffitest  6371  findcard  6372  bj-d0clsepcl  10720  bj-inf2vnlem1  10765