Metamath Proof Explorer
Description: 'Less than' in terms of 'less than or equal to'. (Contributed by Glauco
Siliprandi, 3-Mar-2021)
|
|
Ref |
Expression |
|
Hypotheses |
xrltnled.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℝ* ) |
|
|
xrltnled.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℝ* ) |
|
Assertion |
xrltnled |
⊢ ( 𝜑 → ( 𝐴 < 𝐵 ↔ ¬ 𝐵 ≤ 𝐴 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
xrltnled.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℝ* ) |
2 |
|
xrltnled.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℝ* ) |
3 |
|
xrltnle |
⊢ ( ( 𝐴 ∈ ℝ* ∧ 𝐵 ∈ ℝ* ) → ( 𝐴 < 𝐵 ↔ ¬ 𝐵 ≤ 𝐴 ) ) |
4 |
1 2 3
|
syl2anc |
⊢ ( 𝜑 → ( 𝐴 < 𝐵 ↔ ¬ 𝐵 ≤ 𝐴 ) ) |