Theorem rexrab2 2898

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 2458	. . 3 ⊢ {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}
2	1	rexeqi 2671	. 2 ⊢ (∃𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑}𝜓 ↔ ∃𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}𝜓)
3		ralab2.1	. . 3 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))
4	3	rexab2 2897	. 2 ⊢ (∃𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}𝜓 ↔ ∃𝑦((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒))
5		anass 399	. . . 4 ⊢ (((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ (𝑦 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
6	5	exbii 1599	. . 3 ⊢ (∃𝑦((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ ∃𝑦(𝑦 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
7		df-rex 2455	. . 3 ⊢ (∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒) ↔ ∃𝑦(𝑦 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
8	6, 7	bitr4i 186	. 2 ⊢ (∃𝑦((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒))
9	2, 4, 8	3bitri 205	1 ⊢ (∃𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑}𝜓 ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  ∃wex 1486   ∈ wcel 2142  {cab 2157  ∃wrex 2450  {crab 2453

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 705  ax-5 1441  ax-7 1442  ax-gen 1443  ax-ie1 1487  ax-ie2 1488  ax-8 1498  ax-10 1499  ax-11 1500  ax-i12 1501  ax-bndl 1503  ax-4 1504  ax-17 1520  ax-i9 1524  ax-ial 1528  ax-i5r 1529  ax-ext 2153

This theorem depends on definitions:  df-bi 116  df-tru 1352  df-nf 1455  df-sb 1757  df-clab 2158  df-cleq 2164  df-clel 2167  df-nfc 2302  df-rex 2455  df-rab 2458

This theorem is referenced by: (None)