Description: A trichotomy law for ordinals. (Contributed by NM, 1-Nov-2003) (Proof shortened by Andrew Salmon, 25-Jul-2011)
Ref | Expression | ||
---|---|---|---|
Assertion | ordtri4 | ⊢ ( ( Ord 𝐴 ∧ Ord 𝐵 ) → ( 𝐴 = 𝐵 ↔ ( 𝐴 ⊆ 𝐵 ∧ ¬ 𝐴 ∈ 𝐵 ) ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eqss | ⊢ ( 𝐴 = 𝐵 ↔ ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) ) | |
2 | ordtri1 | ⊢ ( ( Ord 𝐵 ∧ Ord 𝐴 ) → ( 𝐵 ⊆ 𝐴 ↔ ¬ 𝐴 ∈ 𝐵 ) ) | |
3 | 2 | ancoms | ⊢ ( ( Ord 𝐴 ∧ Ord 𝐵 ) → ( 𝐵 ⊆ 𝐴 ↔ ¬ 𝐴 ∈ 𝐵 ) ) |
4 | 3 | anbi2d | ⊢ ( ( Ord 𝐴 ∧ Ord 𝐵 ) → ( ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) ↔ ( 𝐴 ⊆ 𝐵 ∧ ¬ 𝐴 ∈ 𝐵 ) ) ) |
5 | 1 4 | syl5bb | ⊢ ( ( Ord 𝐴 ∧ Ord 𝐵 ) → ( 𝐴 = 𝐵 ↔ ( 𝐴 ⊆ 𝐵 ∧ ¬ 𝐴 ∈ 𝐵 ) ) ) |