Theorem rabrab 3442

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 3439	. . . 4 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
2	1	anbi1i 636	. . 3 ⊢ ((𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ 𝜓) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝜓))
3		anass 474	. . 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 3421	1 ⊢ {𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

This proof depends on syntax axioms:   ∧ wa 401   = wceq 1570   ∈ wcel 2146  {crab 3418

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-12 2216  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419

This theorem is used by:  extwwlkfab  30774  fpwrelmapffs  33149