Theorem ceqsexgv 3524

Description: Elimination of an existential quantifier, using implicit substitution. (Contributed by NM, 29-Dec-1996.)

Hypothesis

Ref	Expression
ceqsexgv.1	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
ceqsexgv	⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

Distinct variable groups:   𝑥,𝐴   𝜓,𝑥

Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)

Proof of Theorem ceqsexgv

Step	Hyp	Ref	Expression
1		nfv 2010	. 2 ⊢ Ⅎ𝑥𝜓
2		ceqsexgv.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
3	1, 2	ceqsexg 3523	1 ⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 385   = wceq 1653  ∃wex 1875   ∈ wcel 2157

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-12 2213  ax-ext 2777

This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2786  df-cleq 2792  df-clel 2795  df-v 3387

This theorem is referenced by:  ceqsrexv  3525  clel3g  3530  elxp5  7346  xpsnen  8286  isssc  16794  metuel2  22698  isgrpo  27877  bj-finsumval0  33646  ismgmOLD  34136  brxrn  34630  pmapjat1  35874  dfatdmfcoafv2  42108