Theorem rexrab2 3695

Description: Existential quantification over a class abstraction. (Contributed by Mario Carneiro, 3-Sep-2015.)

Hypothesis

Ref	Expression
ralab2.1	⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
rexrab2	⊢ (∃𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑}𝜓 ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒))

Distinct variable groups:   𝑥,𝑦   𝑥,𝐴   𝜒,𝑥   𝜑,𝑥   𝜓,𝑦

Allowed substitution hints:   𝜑(𝑦)   𝜓(𝑥)   𝜒(𝑦)   𝐴(𝑦)

Proof of Theorem rexrab2

Step	Hyp	Ref	Expression
1		df-rab 3433	. . 3 ⊢ {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}
2	1	rexeqi 3324	. 2 ⊢ (∃𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑}𝜓 ↔ ∃𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}𝜓)
3		ralab2.1	. . 3 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))
4	3	rexab2 3694	. 2 ⊢ (∃𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}𝜓 ↔ ∃𝑦((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒))
5		anass 469	. . . 4 ⊢ (((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ (𝑦 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
6	5	exbii 1850	. . 3 ⊢ (∃𝑦((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ ∃𝑦(𝑦 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
7		df-rex 3071	. . 3 ⊢ (∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒) ↔ ∃𝑦(𝑦 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
8	6, 7	bitr4i 277	. 2 ⊢ (∃𝑦((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒))
9	2, 4, 8	3bitri 296	1 ⊢ (∃𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑}𝜓 ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396  ∃wex 1781   ∈ wcel 2106  {cab 2709  ∃wrex 3070  {crab 3432

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2703

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-tru 1544  df-ex 1782  df-nf 1786  df-sb 2068  df-clab 2710  df-cleq 2724  df-rex 3071  df-rab 3433

This theorem is referenced by:  frminex  5655  sstotbnd3  36632