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Metamath Proof Explorer

Theorem inxpss

Description: Two ways to say that an intersection with a Cartesian product is a subclass. (Contributed by Peter Mazsa, 16-Jul-2019)

Ref Expression
Assertion inxpss ( ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) ⊆ 𝑆 ↔ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) )

Proof

Step Hyp Ref Expression
1 brinxp2 ⊢ ( 𝑥 ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑥 𝑅 𝑦 ) )
2 1 imbi1i ⊢ ( ( 𝑥 ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 → 𝑥 𝑆 𝑦 ) ↔ ( ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑥 𝑅 𝑦 ) → 𝑥 𝑆 𝑦 ) )
3 impexp ⊢ ( ( ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑥 𝑅 𝑦 ) → 𝑥 𝑆 𝑦 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) ) )
4 2 3 bitri ⊢ ( ( 𝑥 ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 → 𝑥 𝑆 𝑦 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) ) )
5 4 2albii ⊢ ( ∀ 𝑥 ∀ 𝑦 ( 𝑥 ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 → 𝑥 𝑆 𝑦 ) ↔ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) ) )
6 relinxp ⊢ Rel ( 𝑅 ∩ ( 𝐴 × 𝐵 ) )
7 ssrel3 ⊢ ( Rel ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) → ( ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) ⊆ 𝑆 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 → 𝑥 𝑆 𝑦 ) ) )
8 6 7 ax-mp ⊢ ( ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) ⊆ 𝑆 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 → 𝑥 𝑆 𝑦 ) )
9 r2al ⊢ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) ↔ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) ) )
10 5 8 9 3bitr4i ⊢ ( ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) ⊆ 𝑆 ↔ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) )