Metamath Proof Explorer


Theorem ordtri2

Description: A trichotomy law for ordinals. (Contributed by NM, 25-Nov-1995)

Ref Expression
Assertion ordtri2 ( ( Ord 𝐴 ∧ Ord 𝐵 ) → ( 𝐴𝐵 ↔ ¬ ( 𝐴 = 𝐵𝐵𝐴 ) ) )

Proof

Step Hyp Ref Expression
1 ordsseleq ( ( Ord 𝐵 ∧ Ord 𝐴 ) → ( 𝐵𝐴 ↔ ( 𝐵𝐴𝐵 = 𝐴 ) ) )
2 eqcom ( 𝐵 = 𝐴𝐴 = 𝐵 )
3 2 orbi2i ( ( 𝐵𝐴𝐵 = 𝐴 ) ↔ ( 𝐵𝐴𝐴 = 𝐵 ) )
4 orcom ( ( 𝐵𝐴𝐴 = 𝐵 ) ↔ ( 𝐴 = 𝐵𝐵𝐴 ) )
5 3 4 bitri ( ( 𝐵𝐴𝐵 = 𝐴 ) ↔ ( 𝐴 = 𝐵𝐵𝐴 ) )
6 1 5 syl6bb ( ( Ord 𝐵 ∧ Ord 𝐴 ) → ( 𝐵𝐴 ↔ ( 𝐴 = 𝐵𝐵𝐴 ) ) )
7 ordtri1 ( ( Ord 𝐵 ∧ Ord 𝐴 ) → ( 𝐵𝐴 ↔ ¬ 𝐴𝐵 ) )
8 6 7 bitr3d ( ( Ord 𝐵 ∧ Ord 𝐴 ) → ( ( 𝐴 = 𝐵𝐵𝐴 ) ↔ ¬ 𝐴𝐵 ) )
9 8 ancoms ( ( Ord 𝐴 ∧ Ord 𝐵 ) → ( ( 𝐴 = 𝐵𝐵𝐴 ) ↔ ¬ 𝐴𝐵 ) )
10 9 con2bid ( ( Ord 𝐴 ∧ Ord 𝐵 ) → ( 𝐴𝐵 ↔ ¬ ( 𝐴 = 𝐵𝐵𝐴 ) ) )