Theorem ceqsrexv 2956

Description: Elimination of a restricted existential quantifier, using implicit substitution. (Contributed by NM, 30-Apr-2004.)

Hypothesis

Ref	Expression
ceqsrexv.1	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
ceqsrexv	⊢ (𝐴 ∈ 𝐵 → (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜓,𝑥

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem ceqsrexv

Step	Hyp	Ref	Expression
1		df-rex 2534	. . 3 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
2		an12 567	. . . 4 ⊢ ((𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
3	2	exbii 1658	. . 3 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
4	1, 3	bitr4i 187	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)))
5		eleq1 2301	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
6		ceqsrexv.1	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
7	5, 6	anbi12d 477	. . . 4 ⊢ (𝑥 = 𝐴 → ((𝑥 ∈ 𝐵 ∧ 𝜑) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓)))
8	7	ceqsexgv 2955	. . 3 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓)))
9	8	bianabs 619	. 2 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ 𝜓))
10	4, 9	bitrid 192	1 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

This proof depends on syntax axioms:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1402  ∃wex 1545   ∈ wcel 2209  ∃wrex 2529

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-rex 2534  df-v 2823

This theorem is used by:  ceqsrexbv  2957  ceqsrex2v  2958  f1oiso  6032  creur  9292  creui  9293