Theorem ceqsexgv 3611

Description: Elimination of an existential quantifier, using implicit substitution. (Contributed by NM, 29-Dec-1996.) Drop ax-10 2180 and ax-12 2216. (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 401   = wceq 1570  ∃wex 1812   ∈ wcel 2147

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-clel 2839

This theorem is used by:  ceqsrexv  3612  clel3g  3618  elxp5  7956  xpsnen  9105  isssc  18021  metuel2  24911  isgrpo  31126  bj-finsumval0  38397  ismgmOLD  38975  brxrn  39506  pmapjat1  41101  dfatdmfcoafv2  48527