Metamath Proof Explorer


Theorem xpider

Description: A Cartesian square is an equivalence relation (in general, it is not a poset). (Contributed by FL, 31-Jul-2009) (Revised by Mario Carneiro, 12-Aug-2015)

Ref Expression
Assertion xpider ⊢ A × A Er A

Proof

Step Hyp Ref Expression
1 relxp ⊢ Rel ⁡ A × A
2 dmxpid ⊢ dom ⁡ A × A = A
3 cnvxp ⊢ A × A -1 = A × A
4 xpidtr ⊢ A × A ∘ A × A ⊆ A × A
5 uneq1 ⊢ A × A -1 = A × A → A × A -1 ∪ A × A = A × A ∪ A × A
6 unss2 ⊢ A × A ∘ A × A ⊆ A × A → A × A -1 ∪ A × A ∘ A × A ⊆ A × A -1 ∪ A × A
7 unidm ⊢ A × A ∪ A × A = A × A
8 eqtr ⊢ A × A -1 ∪ A × A = A × A ∪ A × A ∧ A × A ∪ A × A = A × A → A × A -1 ∪ A × A = A × A
9 sseq2 ⊢ A × A -1 ∪ A × A = A × A → A × A -1 ∪ A × A ∘ A × A ⊆ A × A -1 ∪ A × A ↔ A × A -1 ∪ A × A ∘ A × A ⊆ A × A
10 9 biimpd ⊢ A × A -1 ∪ A × A = A × A → A × A -1 ∪ A × A ∘ A × A ⊆ A × A -1 ∪ A × A → A × A -1 ∪ A × A ∘ A × A ⊆ A × A
11 8 10 syl ⊢ A × A -1 ∪ A × A = A × A ∪ A × A ∧ A × A ∪ A × A = A × A → A × A -1 ∪ A × A ∘ A × A ⊆ A × A -1 ∪ A × A → A × A -1 ∪ A × A ∘ A × A ⊆ A × A
12 7 11 mpan2 ⊢ A × A -1 ∪ A × A = A × A ∪ A × A → A × A -1 ∪ A × A ∘ A × A ⊆ A × A -1 ∪ A × A → A × A -1 ∪ A × A ∘ A × A ⊆ A × A
13 5 6 12 syl2im ⊢ A × A -1 = A × A → A × A ∘ A × A ⊆ A × A → A × A -1 ∪ A × A ∘ A × A ⊆ A × A
14 3 4 13 mp2 ⊢ A × A -1 ∪ A × A ∘ A × A ⊆ A × A
15 df-er ⊢ A × A Er A ↔ Rel ⁡ A × A ∧ dom ⁡ A × A = A ∧ A × A -1 ∪ A × A ∘ A × A ⊆ A × A
16 1 2 14 15 mpbir3an ⊢ A × A Er A