Theorem ceqsexgv 3485

Description: Elimination of an existential quantifier, using implicit substitution. (Contributed by NM, 29-Dec-1996.)

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		nfv 1995	. 2 ⊢ Ⅎ𝑥𝜓
2		ceqsexgv.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
3	1, 2	ceqsexg 3484	1 ⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 382   = wceq 1631  ∃wex 1852   ∈ wcel 2145

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-9 2154  ax-10 2174  ax-12 2203  ax-ext 2751

This theorem depends on definitions:  df-bi 197  df-an 383  df-or 837  df-tru 1634  df-ex 1853  df-nf 1858  df-sb 2050  df-clab 2758  df-cleq 2764  df-clel 2767  df-v 3353

This theorem is referenced by:  ceqsrexv  3486  clel3g  3491  elxp5  7258  xpsnen  8200  isssc  16687  metuel2  22590  isgrpo  27691  bj-finsumval0  33484  ismgmOLD  33981  brxrn  34478  pmapjat1  35661