Theorem rexab2 3659

Description: Existential quantification over a class abstraction. (Contributed by Mario Carneiro, 3-Sep-2015.) Drop ax-8 2148. (Revised by GG, 1-Dec-2023.)

Hypothesis

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

Assertion

Ref	Expression
rexab2	⊢ (∃𝑥 ∈ {𝑦 ∣ 𝜑}𝜓 ↔ ∃𝑦(𝜑 ∧ 𝜒))

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

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

Proof of Theorem rexab2

Step	Hyp	Ref	Expression
1		df-rex 3090	. 2 ⊢ (∃𝑥 ∈ {𝑦 ∣ 𝜑}𝜓 ↔ ∃𝑥(𝑥 ∈ {𝑦 ∣ 𝜑} ∧ 𝜓))
2		nfsab1 2749	. . . 4 ⊢ Ⅎ𝑦 𝑥 ∈ {𝑦 ∣ 𝜑}
3		nfv 1948	. . . 4 ⊢ Ⅎ𝑦𝜓
4	2, 3	nfan 1933	. . 3 ⊢ Ⅎ𝑦(𝑥 ∈ {𝑦 ∣ 𝜑} ∧ 𝜓)
5		nfv 1948	. . 3 ⊢ Ⅎ𝑥(𝜑 ∧ 𝜒)
6		eleq1ab 2743	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ {𝑦 ∣ 𝜑} ↔ 𝑦 ∈ {𝑦 ∣ 𝜑}))
7		abid 2745	. . . . 5 ⊢ (𝑦 ∈ {𝑦 ∣ 𝜑} ↔ 𝜑)
8	6, 7	bitrdi 290	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ {𝑦 ∣ 𝜑} ↔ 𝜑))
9		ralab2.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))
10	8, 9	anbi12d 644	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ {𝑦 ∣ 𝜑} ∧ 𝜓) ↔ (𝜑 ∧ 𝜒)))
11	4, 5, 10	cbvexv1 2374	. 2 ⊢ (∃𝑥(𝑥 ∈ {𝑦 ∣ 𝜑} ∧ 𝜓) ↔ ∃𝑦(𝜑 ∧ 𝜒))
12	1, 11	bitri 278	1 ⊢ (∃𝑥 ∈ {𝑦 ∣ 𝜑}𝜓 ↔ ∃𝑦(𝜑 ∧ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∃wex 1811   ∈ wcel 2146  {cab 2741  ∃wrex 3089

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-10 2179  ax-11 2196  ax-12 2215

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2101  df-clab 2742  df-rex 3090

This theorem is used by:  rexrab2  3660  tmdgsum2  24427  clrellem  44762  brtrclfv2  44867