Theorem ceqsexgv 3612

Description: Elimination of an existential quantifier, using implicit substitution. (Contributed by NM, 29-Dec-1996.) Drop ax-10 2175 and ax-12 2212. (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 23	. 2 ⊢ (𝑥 = 𝐴 → 𝑥 = 𝐴)
2		ceqsexgv.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
3	1, 2	cgsexg 3498	1 ⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1569  ∃wex 1808   ∈ wcel 2142

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-clel 2837

This theorem is used by:  ceqsrexv  3613  clel3g  3619  elxp5  7918  xpsnen  9047  isssc  17883  metuel2  24733  isgrpo  30860  bj-finsumval0  37957  ismgmOLD  38529  brxrn  39060  pmapjat1  40655  dfatdmfcoafv2  48019