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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 abbii ⊢ { 𝑥 ∣ ( 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∧ 𝜓 ) } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) }
6 df-rab ⊢ { 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∣ 𝜓 } = { 𝑥 ∣ ( 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∧ 𝜓 ) }
7 df-rab ⊢ { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∧ 𝜓 ) } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) }
8 5 6 7 3eqtr4i ⊢ { 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∣ 𝜓 } = { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∧ 𝜓 ) }