Theorem ceqsexgv 3615

Description: Elimination of an existential quantifier, using implicit substitution. (Contributed by NM, 29-Dec-1996.) Drop ax-10 2177 and ax-12 2214. (Revised by GG, 1-Dec-2023.)

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		id 22	. 2 ⊢ (𝑥 = 𝐴 → 𝑥 = 𝐴)
2		ceqsexgv.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
3	1, 2	cgsexg 3500	1 ⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 399   = wceq 1562  ∃wex 1801   ∈ wcel 2144

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1817  ax-4 1831  ax-5 1932  ax-6 1989  ax-7 2030  ax-8 2146

This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1565  df-ex 1802  df-sb 2093  df-clab 2743  df-clel 2839

This theorem is referenced by:  ceqsrexv  3616  clel3g  3622  elxp5  7906  xpsnen  9035  isssc  17855  metuel2  24627  isgrpo  30702  bj-finsumval0  37782  ismgmOLD  38354  brxrn  38887  pmapjat1  40482  dfatdmfcoafv2  47853