Metamath Proof Explorer


Theorem incom

Description: Commutative law for intersection of classes. Exercise 7 of TakeutiZaring p. 17. (Contributed by NM, 21-Jun-1993) (Proof shortened by SN, 12-Dec-2023)

Ref Expression
Assertion incom ( 𝐴 ∩ 𝐵 ) = ( 𝐵 ∩ 𝐴 )

Proof

Step Hyp Ref Expression
1 rabswap ⊢ { 𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵 } = { 𝑥 ∈ 𝐵 ∣ 𝑥 ∈ 𝐴 }
2 dfin5 ⊢ ( 𝐴 ∩ 𝐵 ) = { 𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵 }
3 dfin5 ⊢ ( 𝐵 ∩ 𝐴 ) = { 𝑥 ∈ 𝐵 ∣ 𝑥 ∈ 𝐴 }
4 1 2 3 3eqtr4i ⊢ ( 𝐴 ∩ 𝐵 ) = ( 𝐵 ∩ 𝐴 )