Theorem cgsexg 3498

Description: Implicit substitution inference for general classes. (Contributed by NM, 26-Aug-2007.)

Hypotheses

Ref	Expression
cgsexg.1	⊢ (𝑥 = 𝐴 → 𝜒)
cgsexg.2	⊢ (𝜒 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
cgsexg	⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝜒 ∧ 𝜑) ↔ 𝜓))

Distinct variable groups:   𝑥,𝐴   𝜓,𝑥

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

Proof of Theorem cgsexg

Step	Hyp	Ref	Expression
1		cgsexg.2	. . . 4 ⊢ (𝜒 → (𝜑 ↔ 𝜓))
2	1	biimpa 481	. . 3 ⊢ ((𝜒 ∧ 𝜑) → 𝜓)
3	2	exlimiv 1959	. 2 ⊢ (∃𝑥(𝜒 ∧ 𝜑) → 𝜓)
4		elisset 2844	. . . 4 ⊢ (𝐴 ∈ 𝑉 → ∃𝑥 𝑥 = 𝐴)
5		cgsexg.1	. . . . 5 ⊢ (𝑥 = 𝐴 → 𝜒)
6	5	eximi 1864	. . . 4 ⊢ (∃𝑥 𝑥 = 𝐴 → ∃𝑥𝜒)
7	4, 6	syl 18	. . 3 ⊢ (𝐴 ∈ 𝑉 → ∃𝑥𝜒)
8	1	biimprcd 253	. . . . 5 ⊢ (𝜓 → (𝜒 → 𝜑))
9	8	ancld 559	. . . 4 ⊢ (𝜓 → (𝜒 → (𝜒 ∧ 𝜑)))
10	9	eximdv 1946	. . 3 ⊢ (𝜓 → (∃𝑥𝜒 → ∃𝑥(𝜒 ∧ 𝜑)))
11	7, 10	syl5com 32	. 2 ⊢ (𝐴 ∈ 𝑉 → (𝜓 → ∃𝑥(𝜒 ∧ 𝜑)))
12	3, 11	impbid2 229	1 ⊢ (𝐴 ∈ 𝑉 → (∃𝑥(𝜒 ∧ 𝜑) ↔ 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1569  ∃wex 1808   ∈ wcel 2142

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-clel 2837

This theorem is used by:  ceqsexgv  3612