Theorem ceqsexgv 3593

Description: Elimination of an existential quantifier, using implicit substitution. (Contributed by NM, 29-Dec-1996.) Drop ax-10 2154 and ax-12 2191. (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 3477	1 ⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 397   = wceq 1548  ∃wex 1787   ∈ wcel 2121

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1975  ax-7 2016  ax-8 2123

This theorem depends on definitions:  df-bi 209  df-an 398  df-tru 1551  df-ex 1788  df-sb 2075  df-clab 2720  df-clel 2816

This theorem is referenced by:  ceqsrexv  3594  clel3g  3600  elxp5  7867  xpsnen  8993  isssc  17782  metuel2  24551  isgrpo  30588  bj-finsumval0  37658  ismgmOLD  38230  brxrn  38763  pmapjat1  40358  dfatdmfcoafv2  47729