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Metamath Proof Explorer

Theorem inrab2

Description: Intersection with a restricted class abstraction. (Contributed by NM, 19-Nov-2007)

Ref Expression
Assertion inrab2 ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ 𝐵 ) = { 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∣ 𝜑 }

Proof

Step Hyp Ref Expression
1 df-rab ⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
2 abid1 ⊢ 𝐵 = { 𝑥 ∣ 𝑥 ∈ 𝐵 }
3 1 2 ineq12i ⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ 𝐵 ) = ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ 𝑥 ∈ 𝐵 } )
4 df-rab ⊢ { 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∧ 𝜑 ) }
5 inab ⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ 𝑥 ∈ 𝐵 } ) = { 𝑥 ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝑥 ∈ 𝐵 ) }
6 elin ⊢ ( 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) )
7 6 anbi1i ⊢ ( ( 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∧ 𝜑 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) ∧ 𝜑 ) )
8 an32 ⊢ ( ( ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) ∧ 𝜑 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝑥 ∈ 𝐵 ) )
9 7 8 bitri ⊢ ( ( 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∧ 𝜑 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝑥 ∈ 𝐵 ) )
10 9 abbii ⊢ { 𝑥 ∣ ( 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∧ 𝜑 ) } = { 𝑥 ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝑥 ∈ 𝐵 ) }
11 5 10 eqtr4i ⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ 𝑥 ∈ 𝐵 } ) = { 𝑥 ∣ ( 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∧ 𝜑 ) }
12 4 11 eqtr4i ⊢ { 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∣ 𝜑 } = ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ 𝑥 ∈ 𝐵 } )
13 3 12 eqtr4i ⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ 𝐵 ) = { 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∣ 𝜑 }