Metamath Proof Explorer
Description: Deduction form of dvdsmultr1 . (Contributed by Stanislas Polu, 9-Mar-2020)
|
|
Ref |
Expression |
|
Hypotheses |
dvdsmultr1d.1 |
⊢ ( 𝜑 → 𝐾 ∈ ℤ ) |
|
|
dvdsmultr1d.2 |
⊢ ( 𝜑 → 𝑀 ∈ ℤ ) |
|
|
dvdsmultr1d.3 |
⊢ ( 𝜑 → 𝑁 ∈ ℤ ) |
|
|
dvdsmultr1d.4 |
⊢ ( 𝜑 → 𝐾 ∥ 𝑀 ) |
|
Assertion |
dvdsmultr1d |
⊢ ( 𝜑 → 𝐾 ∥ ( 𝑀 · 𝑁 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
dvdsmultr1d.1 |
⊢ ( 𝜑 → 𝐾 ∈ ℤ ) |
2 |
|
dvdsmultr1d.2 |
⊢ ( 𝜑 → 𝑀 ∈ ℤ ) |
3 |
|
dvdsmultr1d.3 |
⊢ ( 𝜑 → 𝑁 ∈ ℤ ) |
4 |
|
dvdsmultr1d.4 |
⊢ ( 𝜑 → 𝐾 ∥ 𝑀 ) |
5 |
|
dvdsmultr1 |
⊢ ( ( 𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝐾 ∥ 𝑀 → 𝐾 ∥ ( 𝑀 · 𝑁 ) ) ) |
6 |
1 2 3 5
|
syl3anc |
⊢ ( 𝜑 → ( 𝐾 ∥ 𝑀 → 𝐾 ∥ ( 𝑀 · 𝑁 ) ) ) |
7 |
4 6
|
mpd |
⊢ ( 𝜑 → 𝐾 ∥ ( 𝑀 · 𝑁 ) ) |