Theorem rabrabi 3433

Description: Abstract builder restricted to another restricted abstract builder with implicit substitution. (Contributed by AV, 2-Aug-2022.) Avoid ax-10 2178, ax-11 2194 and ax-12 2215. (Revised by GG, 12-Oct-2024.)

Hypothesis

Ref	Expression
rabrabi.1	⊢ (𝑥 = 𝑦 → (𝜒 ↔ 𝜑))

Assertion

Ref	Expression
rabrabi	⊢ {𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ (𝜒 ∧ 𝜓)}

Distinct variable groups:   𝑥,𝑦   𝑦,𝐴   𝜒,𝑦

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

Proof of Theorem rabrabi

Step	Hyp	Ref	Expression
1		df-rab 3415	. . . . . 6 ⊢ {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}
2	1	eleq2i 2854	. . . . 5 ⊢ (𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ↔ 𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)})
3		df-clab 2741	. . . . 5 ⊢ (𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)} ↔ [𝑥 / 𝑦](𝑦 ∈ 𝐴 ∧ 𝜑))
4		eleq1w 2845	. . . . . . 7 ⊢ (𝑦 = 𝑥 → (𝑦 ∈ 𝐴 ↔ 𝑥 ∈ 𝐴))
5		rabrabi.1	. . . . . . . . 9 ⊢ (𝑥 = 𝑦 → (𝜒 ↔ 𝜑))
6	5	bicomd 226	. . . . . . . 8 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜒))
7	6	equcoms 2053	. . . . . . 7 ⊢ (𝑦 = 𝑥 → (𝜑 ↔ 𝜒))
8	4, 7	anbi12d 644	. . . . . 6 ⊢ (𝑦 = 𝑥 → ((𝑦 ∈ 𝐴 ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒)))
9	8	sbievw 2131	. . . . 5 ⊢ ([𝑥 / 𝑦](𝑦 ∈ 𝐴 ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒))
10	2, 3, 9	3bitri 300	. . . 4 ⊢ (𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ↔ (𝑥 ∈ 𝐴 ∧ 𝜒))
11	10	anbi1i 636	. . 3 ⊢ ((𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ∧ 𝜓) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜒) ∧ 𝜓))
12		anass 474	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜒) ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ (𝜒 ∧ 𝜓)))
13	11, 12	bitri 278	. 2 ⊢ ((𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ (𝜒 ∧ 𝜓)))
14	13	rabbia2 3417	1 ⊢ {𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ (𝜒 ∧ 𝜓)}

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570  [wsb 2099   ∈ wcel 2145  {cab 2740  {crab 3414

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-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415

This theorem is used by:  wlksnwwlknvbij  30360