Theorem rabrab 3415

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 3412	. . . 4 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
2	1	anbi1i 630	. . 3 ⊢ ((𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ 𝜓) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝜓))
3		anass 469	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓)))
4	2, 3	bitri 276	. 2 ⊢ ((𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓)))
5	4	rabbia2 3394	1 ⊢ {𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 396   = wceq 1547   ∈ wcel 2119  {crab 3391

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-12 2189  ax-ext 2711

This theorem depends on definitions:  df-bi 208  df-an 397  df-tru 1550  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-rab 3392

This theorem is referenced by:  extwwlkfab  30440  fpwrelmapffs  32826