Theorem ceqsexgv 3595

Description: Elimination of an existential quantifier, using implicit substitution. (Contributed by NM, 29-Dec-1996.) Drop ax-10 2142 and ax-12 2175. (Revised by Gino Giotto, 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 3484	1 ⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399   = wceq 1538  ∃wex 1781   ∈ wcel 2111

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782  df-cleq 2791  df-clel 2870

This theorem is referenced by:  ceqsrexv  3597  clel3g  3602  elxp5  7610  xpsnen  8584  isssc  17082  metuel2  23172  isgrpo  28280  bj-finsumval0  34700  ismgmOLD  35288  brxrn  35786  pmapjat1  37149  dfatdmfcoafv2  43810