Theorem rexrab2 3661

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 3417	. . 3 ⊢ {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}
2	1	rexeqi 3322	. 2 ⊢ (∃𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑}𝜓 ↔ ∃𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}𝜓)
3		ralab2.1	. . 3 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))
4	3	rexab2 3660	. 2 ⊢ (∃𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}𝜓 ↔ ∃𝑦((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒))
5		anass 474	. . . 4 ⊢ (((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ (𝑦 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
6	5	exbii 1881	. . 3 ⊢ (∃𝑦((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ ∃𝑦(𝑦 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
7		df-rex 3091	. . 3 ⊢ (∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒) ↔ ∃𝑦(𝑦 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
8	6, 7	bitr4i 281	. 2 ⊢ (∃𝑦((𝑦 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒))
9	2, 4, 8	3bitri 300	1 ⊢ (∃𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑}𝜓 ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∃wex 1812   ∈ wcel 2147  {cab 2741  ∃wrex 3090  {crab 3416

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-rex 3091  df-rab 3417

This theorem is used by:  frminex  5635  sstotbnd3  38913