Metamath Proof Explorer
Description: Transitive law deduction for 'less than', 'less than or equal to'.
(Contributed by NM, 9-Jan-2006)
|
|
Ref |
Expression |
|
Hypotheses |
ltd.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℝ ) |
|
|
ltd.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℝ ) |
|
|
letrd.3 |
⊢ ( 𝜑 → 𝐶 ∈ ℝ ) |
|
|
ltletrd.4 |
⊢ ( 𝜑 → 𝐴 < 𝐵 ) |
|
|
ltletrd.5 |
⊢ ( 𝜑 → 𝐵 ≤ 𝐶 ) |
|
Assertion |
ltletrd |
⊢ ( 𝜑 → 𝐴 < 𝐶 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
ltd.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℝ ) |
2 |
|
ltd.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℝ ) |
3 |
|
letrd.3 |
⊢ ( 𝜑 → 𝐶 ∈ ℝ ) |
4 |
|
ltletrd.4 |
⊢ ( 𝜑 → 𝐴 < 𝐵 ) |
5 |
|
ltletrd.5 |
⊢ ( 𝜑 → 𝐵 ≤ 𝐶 ) |
6 |
|
ltletr |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐴 < 𝐵 ∧ 𝐵 ≤ 𝐶 ) → 𝐴 < 𝐶 ) ) |
7 |
1 2 3 6
|
syl3anc |
⊢ ( 𝜑 → ( ( 𝐴 < 𝐵 ∧ 𝐵 ≤ 𝐶 ) → 𝐴 < 𝐶 ) ) |
8 |
4 5 7
|
mp2and |
⊢ ( 𝜑 → 𝐴 < 𝐶 ) |