Theorem ceqsrexv 3648

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 3144	. . 3 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
2		an12 643	. . . 4 ⊢ ((𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
3	2	exbii 1844	. . 3 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
4	1, 3	bitr4i 280	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)))
5		eleq1 2900	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
6		ceqsrexv.1	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
7	5, 6	anbi12d 632	. . . 4 ⊢ (𝑥 = 𝐴 → ((𝑥 ∈ 𝐵 ∧ 𝜑) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓)))
8	7	ceqsexgv 3646	. . 3 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓)))
9	8	bianabs 544	. 2 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ 𝜓))
10	4, 9	syl5bb 285	1 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398   = wceq 1533  ∃wex 1776   ∈ wcel 2110  ∃wrex 3139

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1777  df-cleq 2814  df-clel 2893  df-rex 3144

This theorem is referenced by:  ceqsrexbv  3649  ceqsrex2v  3650  reuxfrd  3738  f1oiso  7098  creur  11626  creui  11627  deg1ldg  24680  ulm2  24967  iscgra1  26590  reuxfrdf  30249  poimirlem24  34910  eqlkr3  36231  diclspsn  38324  rmxdiophlem  39605  expdiophlem1  39611  expdiophlem2  39612