Theorem spcv 2919

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 2912	. 2 ⊢ (𝐴 ∈ V → (∀𝑥𝜑 → 𝜓))
4	1, 3	ax-mp 5	1 ⊢ (∀𝑥𝜑 → 𝜓)

This proof depends on syntax axioms:   → wi 4   ↔ wb 105  ∀wal 1400   = wceq 1402   ∈ wcel 2209  Vcvv 2821

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823

This theorem is used by:  morex  3010  exmidexmid  4333  exmidsssn  4339  exmidel  4342  rext  4355  ontr2exmid  4672  regexmidlem1  4680  reg2exmid  4683  relop  4930  uchoice  6371  disjxp1  6472  rdgtfr  6645  ssfiexmid  7178  ssfiexmidt  7180  domfiexmid  7182  diffitest  7191  findcard  7192  exmidpw2en  7219  fiintim  7238  fisseneq  7242  finomni  7480  exmidomni  7482  exmidlpo  7483  ballotfilem2  13230  exmidunben  13319  ivthreinc  15748  bj-d0clsepcl  16963  bj-inf2vnlem1  17008  subctctexmid  17042  wexmiddiffilem  17055  wexmiddifxy  17058