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 ⊢ ≤ ˙ = ≤ K ∩ B × B
Assertion prsss ⊢ K ∈ Proset ∧ A ⊆ B → ≤ ˙ ∩ A × A = ≤ K ∩ A × A

Proof

Step Hyp Ref Expression
1 ordtNEW.b ⊢ B = Base K
2 ordtNEW.l ⊢ ≤ ˙ = ≤ K ∩ B × B
3 2 ineq1i ⊢ ≤ ˙ ∩ A × A = ≤ K ∩ B × B ∩ A × A
4 inass ⊢ ≤ K ∩ B × B ∩ A × A = ≤ K ∩ B × B ∩ A × A
5 3 4 eqtri ⊢ ≤ ˙ ∩ A × A = ≤ 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 → ≤ K ∩ B × B ∩ A × A = ≤ K ∩ A × A
11 5 10 eqtrid ⊢ A ⊆ B → ≤ ˙ ∩ A × A = ≤ K ∩ A × A
12 11 adantl ⊢ K ∈ Proset ∧ A ⊆ B → ≤ ˙ ∩ A × A = ≤ K ∩ A × A