Theorem ceqsexgv 3623

Description: Elimination of an existential quantifier, using implicit substitution. (Contributed by NM, 29-Dec-1996.) Drop ax-10 2142 and ax-12 2178. (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 3495	1 ⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1540  ∃wex 1779   ∈ wcel 2109

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2709  df-clel 2804

This theorem is referenced by:  ceqsrexv  3624  clel3g  3630  elxp5  7902  xpsnen  9029  isssc  17789  metuel2  24460  isgrpo  30433  bj-finsumval0  37280  ismgmOLD  37851  brxrn  38363  pmapjat1  39854  dfatdmfcoafv2  47259