Metamath Proof Explorer
Description: Ordering on reals satisfies strict trichotomy. (Contributed by Mario
Carneiro, 27-May-2016)
|
|
Ref |
Expression |
|
Hypotheses |
ltd.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℝ ) |
|
|
ltd.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℝ ) |
|
Assertion |
lttrid |
⊢ ( 𝜑 → ( 𝐴 < 𝐵 ↔ ¬ ( 𝐴 = 𝐵 ∨ 𝐵 < 𝐴 ) ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
ltd.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℝ ) |
2 |
|
ltd.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℝ ) |
3 |
|
axlttri |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ ¬ ( 𝐴 = 𝐵 ∨ 𝐵 < 𝐴 ) ) ) |
4 |
1 2 3
|
syl2anc |
⊢ ( 𝜑 → ( 𝐴 < 𝐵 ↔ ¬ ( 𝐴 = 𝐵 ∨ 𝐵 < 𝐴 ) ) ) |