Metamath Proof Explorer


Theorem prsss

Description: Relation of a subproset. (Contributed by Thierry Arnoux, 13-Sep-2018)

Ref Expression
Hypotheses ordtNEW.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
ordtNEW.l ⊢ ≤ = ( ( le ‘ 𝐾 ) ∩ ( 𝐵 × 𝐵 ) )
Assertion prsss ( ( 𝐾 ∈ Proset ∧ 𝐴 ⊆ 𝐵 ) → ( ≤ ∩ ( 𝐴 × 𝐴 ) ) = ( ( le ‘ 𝐾 ) ∩ ( 𝐴 × 𝐴 ) ) )

Proof

Step Hyp Ref Expression
1 ordtNEW.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 ordtNEW.l ⊢ ≤ = ( ( le ‘ 𝐾 ) ∩ ( 𝐵 × 𝐵 ) )
3 2 ineq1i ⊢ ( ≤ ∩ ( 𝐴 × 𝐴 ) ) = ( ( ( le ‘ 𝐾 ) ∩ ( 𝐵 × 𝐵 ) ) ∩ ( 𝐴 × 𝐴 ) )
4 inass ⊢ ( ( ( le ‘ 𝐾 ) ∩ ( 𝐵 × 𝐵 ) ) ∩ ( 𝐴 × 𝐴 ) ) = ( ( le ‘ 𝐾 ) ∩ ( ( 𝐵 × 𝐵 ) ∩ ( 𝐴 × 𝐴 ) ) )
5 3 4 eqtri ⊢ ( ≤ ∩ ( 𝐴 × 𝐴 ) ) = ( ( le ‘ 𝐾 ) ∩ ( ( 𝐵 × 𝐵 ) ∩ ( 𝐴 × 𝐴 ) ) )
6 xpss12 ⊢ ( ( 𝐴 ⊆ 𝐵 ∧ 𝐴 ⊆ 𝐵 ) → ( 𝐴 × 𝐴 ) ⊆ ( 𝐵 × 𝐵 ) )
7 6 anidms ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝐴 × 𝐴 ) ⊆ ( 𝐵 × 𝐵 ) )
8 sseqin2 ⊢ ( ( 𝐴 × 𝐴 ) ⊆ ( 𝐵 × 𝐵 ) ↔ ( ( 𝐵 × 𝐵 ) ∩ ( 𝐴 × 𝐴 ) ) = ( 𝐴 × 𝐴 ) )
9 7 8 sylib ⊢ ( 𝐴 ⊆ 𝐵 → ( ( 𝐵 × 𝐵 ) ∩ ( 𝐴 × 𝐴 ) ) = ( 𝐴 × 𝐴 ) )
10 9 ineq2d ⊢ ( 𝐴 ⊆ 𝐵 → ( ( le ‘ 𝐾 ) ∩ ( ( 𝐵 × 𝐵 ) ∩ ( 𝐴 × 𝐴 ) ) ) = ( ( le ‘ 𝐾 ) ∩ ( 𝐴 × 𝐴 ) ) )
11 5 10 eqtrid ⊢ ( 𝐴 ⊆ 𝐵 → ( ≤ ∩ ( 𝐴 × 𝐴 ) ) = ( ( le ‘ 𝐾 ) ∩ ( 𝐴 × 𝐴 ) ) )
12 11 adantl ⊢ ( ( 𝐾 ∈ Proset ∧ 𝐴 ⊆ 𝐵 ) → ( ≤ ∩ ( 𝐴 × 𝐴 ) ) = ( ( le ‘ 𝐾 ) ∩ ( 𝐴 × 𝐴 ) ) )