Theorem rabrab 3447

Description: Abstract builder restricted to another restricted abstract builder. (Contributed by Thierry Arnoux, 30-Aug-2017.)

Assertion

Ref	Expression
rabrab	⊢ {𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

Proof of Theorem rabrab

Step	Hyp	Ref	Expression
1		rabid 3444	. . . 4 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
2	1	anbi1i 635	. . 3 ⊢ ((𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ 𝜓) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝜓))
3		anass 473	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓)))
4	2, 3	bitri 278	. 2 ⊢ ((𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓)))
5	4	rabbia2 3426	1 ⊢ {𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 400   = wceq 1567   ∈ wcel 2149  {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-12 2219  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:  extwwlkfab  30643  fpwrelmapffs  33019