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Metamath Proof Explorer

Theorem inxpss3

Description: Two ways to say that an intersection with a Cartesian product is a subclass (see also inxpss ). (Contributed by Peter Mazsa, 8-Mar-2019)

Ref Expression
Assertion inxpss3 ( ∀ 𝑥 ∀ 𝑦 ( 𝑥 ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 → 𝑥 ( 𝑆 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 ) ↔ ∀ 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 brinxp2 ⊢ ( 𝑥 ( 𝑆 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑥 𝑆 𝑦 ) )
3 1 2 imbi12i ⊢ ( ( 𝑥 ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 → 𝑥 ( 𝑆 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 ) ↔ ( ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑥 𝑅 𝑦 ) → ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑥 𝑆 𝑦 ) ) )
4 imdistan ⊢ ( ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) ) ↔ ( ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑥 𝑅 𝑦 ) → ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑥 𝑆 𝑦 ) ) )
5 3 4 bitr4i ⊢ ( ( 𝑥 ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 → 𝑥 ( 𝑆 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) ) )
6 5 2albii ⊢ ( ∀ 𝑥 ∀ 𝑦 ( 𝑥 ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 → 𝑥 ( 𝑆 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 ) ↔ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) ) )
7 r2al ⊢ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) ↔ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) ) )
8 6 7 bitr4i ⊢ ( ∀ 𝑥 ∀ 𝑦 ( 𝑥 ( 𝑅 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 → 𝑥 ( 𝑆 ∩ ( 𝐴 × 𝐵 ) ) 𝑦 ) ↔ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝑥 𝑅 𝑦 → 𝑥 𝑆 𝑦 ) )