Theorem rabrabi 3453

Description: Abstract builder restricted to another restricted abstract builder with implicit substitution. (Contributed by AV, 2-Aug-2022.) Avoid ax-10 2139, ax-11 2155 and ax-12 2175. (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 3434	. . . . . 6 ⊢ {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)}
2	1	eleq2i 2831	. . . . 5 ⊢ (𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ↔ 𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)})
3		df-clab 2713	. . . . 5 ⊢ (𝑥 ∈ {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝜑)} ↔ [𝑥 / 𝑦](𝑦 ∈ 𝐴 ∧ 𝜑))
4		eleq1w 2822	. . . . . . 7 ⊢ (𝑦 = 𝑥 → (𝑦 ∈ 𝐴 ↔ 𝑥 ∈ 𝐴))
5		rabrabi.1	. . . . . . . . 9 ⊢ (𝑥 = 𝑦 → (𝜒 ↔ 𝜑))
6	5	bicomd 223	. . . . . . . 8 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜒))
7	6	equcoms 2017	. . . . . . 7 ⊢ (𝑦 = 𝑥 → (𝜑 ↔ 𝜒))
8	4, 7	anbi12d 632	. . . . . 6 ⊢ (𝑦 = 𝑥 → ((𝑦 ∈ 𝐴 ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒)))
9	8	sbievw 2091	. . . . 5 ⊢ ([𝑥 / 𝑦](𝑦 ∈ 𝐴 ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒))
10	2, 3, 9	3bitri 297	. . . 4 ⊢ (𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ↔ (𝑥 ∈ 𝐴 ∧ 𝜒))
11	10	anbi1i 624	. . 3 ⊢ ((𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ∧ 𝜓) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜒) ∧ 𝜓))
12		anass 468	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜒) ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ (𝜒 ∧ 𝜓)))
13	11, 12	bitri 275	. 2 ⊢ ((𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ (𝜒 ∧ 𝜓)))
14	13	rabbia2 3436	1 ⊢ {𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜑} ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ (𝜒 ∧ 𝜓)}

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1537  [wsb 2062   ∈ wcel 2106  {cab 2712  {crab 3433

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rab 3434

This theorem is referenced by:  wlksnwwlknvbij  29938