Metamath Proof Explorer
Description: Trichotomy law for 'less than'. (Contributed by NM, 5-May-1999)
|
|
Ref |
Expression |
|
Assertion |
lttri3 |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 = 𝐵 ↔ ( ¬ 𝐴 < 𝐵 ∧ ¬ 𝐵 < 𝐴 ) ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
ltso |
⊢ < Or ℝ |
2 |
|
sotrieq2 |
⊢ ( ( < Or ℝ ∧ ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) ) → ( 𝐴 = 𝐵 ↔ ( ¬ 𝐴 < 𝐵 ∧ ¬ 𝐵 < 𝐴 ) ) ) |
3 |
1 2
|
mpan |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 = 𝐵 ↔ ( ¬ 𝐴 < 𝐵 ∧ ¬ 𝐵 < 𝐴 ) ) ) |