Metamath Proof Explorer


Theorem rabrab

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

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

Proof

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