Metamath Proof Explorer


Theorem brdifun

Description: Evaluate the incomparability relation. (Contributed by Mario Carneiro, 9-Jul-2014)

Ref Expression
Hypothesis swoer.1 ⊢ R = X × X ∖ < ˙ ∪ < ˙ -1
Assertion brdifun ⊢ A ∈ X ∧ B ∈ X → A R B ↔ ¬ A < ˙ B ∨ B < ˙ A

Proof

Step Hyp Ref Expression
1 swoer.1 ⊢ R = X × X ∖ < ˙ ∪ < ˙ -1
2 opelxpi ⊢ A ∈ X ∧ B ∈ X → A B ∈ X × X
3 df-br ⊢ A X × X B ↔ A B ∈ X × X
4 2 3 sylibr ⊢ A ∈ X ∧ B ∈ X → A X × X B
5 1 breqi ⊢ A R B ↔ A X × X ∖ < ˙ ∪ < ˙ -1 B
6 brdif ⊢ A X × X ∖ < ˙ ∪ < ˙ -1 B ↔ A X × X B ∧ ¬ A < ˙ ∪ < ˙ -1 B
7 5 6 bitri ⊢ A R B ↔ A X × X B ∧ ¬ A < ˙ ∪ < ˙ -1 B
8 7 baib ⊢ A X × X B → A R B ↔ ¬ A < ˙ ∪ < ˙ -1 B
9 4 8 syl ⊢ A ∈ X ∧ B ∈ X → A R B ↔ ¬ A < ˙ ∪ < ˙ -1 B
10 brun ⊢ A < ˙ ∪ < ˙ -1 B ↔ A < ˙ B ∨ A < ˙ -1 B
11 brcnvg ⊢ A ∈ X ∧ B ∈ X → A < ˙ -1 B ↔ B < ˙ A
12 11 orbi2d ⊢ A ∈ X ∧ B ∈ X → A < ˙ B ∨ A < ˙ -1 B ↔ A < ˙ B ∨ B < ˙ A
13 10 12 bitrid ⊢ A ∈ X ∧ B ∈ X → A < ˙ ∪ < ˙ -1 B ↔ A < ˙ B ∨ B < ˙ A
14 13 notbid ⊢ A ∈ X ∧ B ∈ X → ¬ A < ˙ ∪ < ˙ -1 B ↔ ¬ A < ˙ B ∨ B < ˙ A
15 9 14 bitrd ⊢ A ∈ X ∧ B ∈ X → A R B ↔ ¬ A < ˙ B ∨ B < ˙ A