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