Theorem ceqsexg 3615

Description: A representation of explicit substitution of a class for a variable, inferred from an implicit substitution hypothesis. (Contributed by NM, 11-Oct-2004.)

Hypotheses

Ref	Expression
ceqsexg.1	⊢ Ⅎ𝑥𝜓
ceqsexg.2	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
ceqsexg	⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

Distinct variable group:   𝑥,𝐴

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥)   𝑉(𝑥)

Proof of Theorem ceqsexg

Step	Hyp	Ref	Expression
1		nfe1 2188	. . 3 ⊢ Ⅎ𝑥∃𝑥(𝑥 = 𝐴 ∧ 𝜑)
2		ceqsexg.1	. . 3 ⊢ Ⅎ𝑥𝜓
3	1, 2	nfbi 1936	. 2 ⊢ Ⅎ𝑥(∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓)
4		ceqex 3614	. . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
5		ceqsexg.2	. . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
6	4, 5	bibi12d 348	. 2 ⊢ (𝑥 = 𝐴 → ((𝜑 ↔ 𝜑) ↔ (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓)))
7		biid 264	. 2 ⊢ (𝜑 ↔ 𝜑)
8	3, 6, 7	vtoclg1f 3538	1 ⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570  ∃wex 1812  Ⅎwnf 1816   ∈ wcel 2146

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 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-10 2179  ax-12 2216  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-v 3460

This theorem is used by: (None)