Metamath Proof Explorer


Theorem bj-xpcossxp

Description: The composition of two Cartesian products is included in the expected Cartesian product. There is equality if ( B i^i C ) =/= (/) , see xpcogend . (Contributed by BJ, 22-May-2024)

Ref Expression
Assertion bj-xpcossxp ( ( 𝐶 × 𝐷 ) ∘ ( 𝐴 × 𝐵 ) ) ⊆ ( 𝐴 × 𝐷 )

Proof

Step Hyp Ref Expression
1 brxp ( 𝑥 ( 𝐴 × 𝐵 ) 𝑦 ↔ ( 𝑥𝐴𝑦𝐵 ) )
2 brxp ( 𝑦 ( 𝐶 × 𝐷 ) 𝑡 ↔ ( 𝑦𝐶𝑡𝐷 ) )
3 1 2 anbi12i ( ( 𝑥 ( 𝐴 × 𝐵 ) 𝑦𝑦 ( 𝐶 × 𝐷 ) 𝑡 ) ↔ ( ( 𝑥𝐴𝑦𝐵 ) ∧ ( 𝑦𝐶𝑡𝐷 ) ) )
4 an43 ( ( ( 𝑥𝐴𝑦𝐵 ) ∧ ( 𝑦𝐶𝑡𝐷 ) ) ↔ ( ( 𝑥𝐴𝑡𝐷 ) ∧ ( 𝑦𝐵𝑦𝐶 ) ) )
5 3 4 bitri ( ( 𝑥 ( 𝐴 × 𝐵 ) 𝑦𝑦 ( 𝐶 × 𝐷 ) 𝑡 ) ↔ ( ( 𝑥𝐴𝑡𝐷 ) ∧ ( 𝑦𝐵𝑦𝐶 ) ) )
6 5 exbii ( ∃ 𝑦 ( 𝑥 ( 𝐴 × 𝐵 ) 𝑦𝑦 ( 𝐶 × 𝐷 ) 𝑡 ) ↔ ∃ 𝑦 ( ( 𝑥𝐴𝑡𝐷 ) ∧ ( 𝑦𝐵𝑦𝐶 ) ) )
7 19.42v ( ∃ 𝑦 ( ( 𝑥𝐴𝑡𝐷 ) ∧ ( 𝑦𝐵𝑦𝐶 ) ) ↔ ( ( 𝑥𝐴𝑡𝐷 ) ∧ ∃ 𝑦 ( 𝑦𝐵𝑦𝐶 ) ) )
8 7 simplbi ( ∃ 𝑦 ( ( 𝑥𝐴𝑡𝐷 ) ∧ ( 𝑦𝐵𝑦𝐶 ) ) → ( 𝑥𝐴𝑡𝐷 ) )
9 6 8 sylbi ( ∃ 𝑦 ( 𝑥 ( 𝐴 × 𝐵 ) 𝑦𝑦 ( 𝐶 × 𝐷 ) 𝑡 ) → ( 𝑥𝐴𝑡𝐷 ) )
10 9 ssopab2i { ⟨ 𝑥 , 𝑡 ⟩ ∣ ∃ 𝑦 ( 𝑥 ( 𝐴 × 𝐵 ) 𝑦𝑦 ( 𝐶 × 𝐷 ) 𝑡 ) } ⊆ { ⟨ 𝑥 , 𝑡 ⟩ ∣ ( 𝑥𝐴𝑡𝐷 ) }
11 df-co ( ( 𝐶 × 𝐷 ) ∘ ( 𝐴 × 𝐵 ) ) = { ⟨ 𝑥 , 𝑡 ⟩ ∣ ∃ 𝑦 ( 𝑥 ( 𝐴 × 𝐵 ) 𝑦𝑦 ( 𝐶 × 𝐷 ) 𝑡 ) }
12 df-xp ( 𝐴 × 𝐷 ) = { ⟨ 𝑥 , 𝑡 ⟩ ∣ ( 𝑥𝐴𝑡𝐷 ) }
13 10 11 12 3sstr4i ( ( 𝐶 × 𝐷 ) ∘ ( 𝐴 × 𝐵 ) ) ⊆ ( 𝐴 × 𝐷 )