Theorem rabrab 3438

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 3435	. . . 4 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
2	1	anbi1i 633	. . 3 ⊢ ((𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ 𝜓) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝜓))
3		anass 472	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓)))
4	2, 3	bitri 277	. 2 ⊢ ((𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓)))
5	4	rabbia2 3417	1 ⊢ {𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 399   = wceq 1560   ∈ wcel 2142  {crab 3414

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-12 2212  ax-ext 2734

This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1563  df-ex 1800  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415

This theorem is referenced by:  extwwlkfab  30554  fpwrelmapffs  32936