Theorem rexrab2 3662

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 3414	. . 3 ⊢ {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}
2	1	rexeqi 3318	. 2 ⊢ (∃𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑}𝜓 ↔ ∃𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}𝜓)
3		ralab2.1	. . 3 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))
4	3	rexab2 3661	. 2 ⊢ (∃𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}𝜓 ↔ ∃𝑦((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒))
5		anass 472	. . . 4 ⊢ (((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ (𝑦 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
6	5	exbii 1867	. . 3 ⊢ (∃𝑦((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ ∃𝑦(𝑦 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
7		df-rex 3086	. . 3 ⊢ (∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒) ↔ ∃𝑦(𝑦 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
8	6, 7	bitr4i 280	. 2 ⊢ (∃𝑦((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒))
9	2, 4, 8	3bitri 299	1 ⊢ (∃𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑}𝜓 ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 399  ∃wex 1798   ∈ wcel 2141  {cab 2739  ∃wrex 3085  {crab 3413

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-9 2151  ax-10 2174  ax-11 2190  ax-12 2211  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1562  df-ex 1799  df-nf 1803  df-sb 2090  df-clab 2740  df-cleq 2753  df-rex 3086  df-rab 3414

This theorem is referenced by:  frminex  5624  sstotbnd3  38239