Metamath Proof Explorer
Description: The divides relation is transitive, a deduction version of dvdstr .
(Contributed by metakunt, 12-May-2024)
|
|
Ref |
Expression |
|
Hypotheses |
dvdstrd.1 |
⊢ ( 𝜑 → 𝐾 ∈ ℤ ) |
|
|
dvdstrd.2 |
⊢ ( 𝜑 → 𝑀 ∈ ℤ ) |
|
|
dvdstrd.3 |
⊢ ( 𝜑 → 𝑁 ∈ ℤ ) |
|
|
dvdstrd.4 |
⊢ ( 𝜑 → 𝐾 ∥ 𝑀 ) |
|
|
dvdstrd.5 |
⊢ ( 𝜑 → 𝑀 ∥ 𝑁 ) |
|
Assertion |
dvdstrd |
⊢ ( 𝜑 → 𝐾 ∥ 𝑁 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
dvdstrd.1 |
⊢ ( 𝜑 → 𝐾 ∈ ℤ ) |
2 |
|
dvdstrd.2 |
⊢ ( 𝜑 → 𝑀 ∈ ℤ ) |
3 |
|
dvdstrd.3 |
⊢ ( 𝜑 → 𝑁 ∈ ℤ ) |
4 |
|
dvdstrd.4 |
⊢ ( 𝜑 → 𝐾 ∥ 𝑀 ) |
5 |
|
dvdstrd.5 |
⊢ ( 𝜑 → 𝑀 ∥ 𝑁 ) |
6 |
|
dvdstr |
⊢ ( ( 𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( ( 𝐾 ∥ 𝑀 ∧ 𝑀 ∥ 𝑁 ) → 𝐾 ∥ 𝑁 ) ) |
7 |
1 2 3 6
|
syl3anc |
⊢ ( 𝜑 → ( ( 𝐾 ∥ 𝑀 ∧ 𝑀 ∥ 𝑁 ) → 𝐾 ∥ 𝑁 ) ) |
8 |
4 5 7
|
mp2and |
⊢ ( 𝜑 → 𝐾 ∥ 𝑁 ) |