Theorem rabrab 3440

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 3437	. . . 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 3419	1 ⊢ {𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 400   = wceq 1570   ∈ wcel 2143  {crab 3416

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-12 2213  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1573  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3417

This theorem is referenced by:  extwwlkfab  30703  fpwrelmapffs  33079