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Metamath Proof Explorer

Theorem inrab

Description: Intersection of two restricted class abstractions. (Contributed by NM, 1-Sep-2006)

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

Proof

Step Hyp Ref Expression
1 df-rab ⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
2 df-rab ⊢ { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) }
3 1 2 ineq12i ⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ { 𝑥 ∈ 𝐴 ∣ 𝜓 } ) = ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } )
4 df-rab ⊢ { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∧ 𝜓 ) } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) }
5 inab ⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } ) = { 𝑥 ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) }
6 anandi ⊢ ( ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) )
7 6 abbii ⊢ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) } = { 𝑥 ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) }
8 5 7 eqtr4i ⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } ) = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) }
9 4 8 eqtr4i ⊢ { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∧ 𝜓 ) } = ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } )
10 3 9 eqtr4i ⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ { 𝑥 ∈ 𝐴 ∣ 𝜓 } ) = { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∧ 𝜓 ) }