Theorem rabrabi 3442

Description: Abstract builder restricted to another restricted abstract builder with implicit substitution. (Contributed by AV, 2-Aug-2022.) Avoid ax-10 2182, ax-11 2198 and ax-12 2219. (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 3424	. . . . . 6 ⊢ {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}
2	1	eleq2i 2861	. . . . 5 ⊢ (𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ↔ 𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)})
3		df-clab 2748	. . . . 5 ⊢ (𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)} ↔ [𝑥 / 𝑦](𝑦 ∈ 𝐴 ∧ 𝜑))
4		eleq1w 2852	. . . . . . 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 2047	. . . . . . 7 ⊢ (𝑦 = 𝑥 → (𝜑 ↔ 𝜒))
8	4, 7	anbi12d 643	. . . . . 6 ⊢ (𝑦 = 𝑥 → ((𝑦 ∈ 𝐴 ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒)))
9	8	sbievw 2134	. . . . 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 635	. . 3 ⊢ ((𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ∧ 𝜓) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜒) ∧ 𝜓))
12		anass 473	. . 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 3426	1 ⊢ {𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ (𝜒 ∧ 𝜓)}

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1567  [wsb 2097   ∈ wcel 2149  {cab 2747  {crab 3423

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1570  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-rab 3424

This theorem is referenced by:  wlksnwwlknvbij  30198