Metamath Proof Explorer


Theorem prsssdm

Description: Domain of a subproset relation. (Contributed by Thierry Arnoux, 12-Sep-2018)

Ref Expression
Hypotheses ordtNEW.b ⊢ B = Base K
ordtNEW.l ⊢ ≤ ˙ = ≤ K ∩ B × B
Assertion prsssdm ⊢ K ∈ Proset ∧ A ⊆ B → dom ⁡ ≤ ˙ ∩ A × A = A

Proof

Step Hyp Ref Expression
1 ordtNEW.b ⊢ B = Base K
2 ordtNEW.l ⊢ ≤ ˙ = ≤ K ∩ B × B
3 1 2 prsss ⊢ K ∈ Proset ∧ A ⊆ B → ≤ ˙ ∩ A × A = ≤ K ∩ A × A
4 3 dmeqd ⊢ K ∈ Proset ∧ A ⊆ B → dom ⁡ ≤ ˙ ∩ A × A = dom ⁡ ≤ K ∩ A × A
5 1 ressprs ⊢ K ∈ Proset ∧ A ⊆ B → K ↾ 𝑠 A ∈ Proset
6 eqid ⊢ Base K ↾ 𝑠 A = Base K ↾ 𝑠 A
7 eqid ⊢ ≤ K ↾ 𝑠 A ∩ Base K ↾ 𝑠 A × Base K ↾ 𝑠 A = ≤ K ↾ 𝑠 A ∩ Base K ↾ 𝑠 A × Base K ↾ 𝑠 A
8 6 7 prsdm ⊢ K ↾ 𝑠 A ∈ Proset → dom ⁡ ≤ K ↾ 𝑠 A ∩ Base K ↾ 𝑠 A × Base K ↾ 𝑠 A = Base K ↾ 𝑠 A
9 5 8 syl ⊢ K ∈ Proset ∧ A ⊆ B → dom ⁡ ≤ K ↾ 𝑠 A ∩ Base K ↾ 𝑠 A × Base K ↾ 𝑠 A = Base K ↾ 𝑠 A
10 eqid ⊢ K ↾ 𝑠 A = K ↾ 𝑠 A
11 10 1 ressbas2 ⊢ A ⊆ B → A = Base K ↾ 𝑠 A
12 fvex ⊢ Base K ↾ 𝑠 A ∈ V
13 11 12 eqeltrdi ⊢ A ⊆ B → A ∈ V
14 eqid ⊢ ≤ K = ≤ K
15 10 14 ressle ⊢ A ∈ V → ≤ K = ≤ K ↾ 𝑠 A
16 13 15 syl ⊢ A ⊆ B → ≤ K = ≤ K ↾ 𝑠 A
17 16 adantl ⊢ K ∈ Proset ∧ A ⊆ B → ≤ K = ≤ K ↾ 𝑠 A
18 11 adantl ⊢ K ∈ Proset ∧ A ⊆ B → A = Base K ↾ 𝑠 A
19 18 sqxpeqd ⊢ K ∈ Proset ∧ A ⊆ B → A × A = Base K ↾ 𝑠 A × Base K ↾ 𝑠 A
20 17 19 ineq12d ⊢ K ∈ Proset ∧ A ⊆ B → ≤ K ∩ A × A = ≤ K ↾ 𝑠 A ∩ Base K ↾ 𝑠 A × Base K ↾ 𝑠 A
21 20 dmeqd ⊢ K ∈ Proset ∧ A ⊆ B → dom ⁡ ≤ K ∩ A × A = dom ⁡ ≤ K ↾ 𝑠 A ∩ Base K ↾ 𝑠 A × Base K ↾ 𝑠 A
22 9 21 18 3eqtr4d ⊢ K ∈ Proset ∧ A ⊆ B → dom ⁡ ≤ K ∩ A × A = A
23 4 22 eqtrd ⊢ K ∈ Proset ∧ A ⊆ B → dom ⁡ ≤ ˙ ∩ A × A = A