Theorem rexab2 3657

Description: Existential quantification over a class abstraction. (Contributed by Mario Carneiro, 3-Sep-2015.) Drop ax-8 2147. (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 3088	. 2 ⊢ (∃𝑥 ∈ {𝑦 ∣ 𝜑}𝜓 ↔ ∃𝑥(𝑥 ∈ {𝑦 ∣ 𝜑} ∧ 𝜓))
2		nfsab1 2747	. . . 4 ⊢ Ⅎ𝑦 𝑥 ∈ {𝑦 ∣ 𝜑}
3		nfv 1947	. . . 4 ⊢ Ⅎ𝑦𝜓
4	2, 3	nfan 1932	. . 3 ⊢ Ⅎ𝑦(𝑥 ∈ {𝑦 ∣ 𝜑} ∧ 𝜓)
5		nfv 1947	. . 3 ⊢ Ⅎ𝑥(𝜑 ∧ 𝜒)
6		eleq1ab 2741	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ {𝑦 ∣ 𝜑} ↔ 𝑦 ∈ {𝑦 ∣ 𝜑}))
7		abid 2743	. . . . 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 2372	. 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 1812   ∈ wcel 2145  {cab 2739  ∃wrex 3087

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 1943  ax-6 2000  ax-7 2041  ax-10 2178  ax-11 2194  ax-12 2213

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 2100  df-clab 2740  df-rex 3088

This theorem is used by:  rexrab2  3658  tmdgsum2  24415  clrellem  44621  brtrclfv2  44726