Metamath Proof Explorer
Description: Deduction form of dvds2sub . (Contributed by Stanislas Polu, 9-Mar-2020)
|
|
Ref |
Expression |
|
Hypotheses |
dvds2subd.k |
⊢ ( 𝜑 → 𝐾 ∈ ℤ ) |
|
|
dvds2subd.m |
⊢ ( 𝜑 → 𝑀 ∈ ℤ ) |
|
|
dvds2subd.n |
⊢ ( 𝜑 → 𝑁 ∈ ℤ ) |
|
|
dvds2subd.1 |
⊢ ( 𝜑 → 𝐾 ∥ 𝑀 ) |
|
|
dvds2subd.2 |
⊢ ( 𝜑 → 𝐾 ∥ 𝑁 ) |
|
Assertion |
dvds2subd |
⊢ ( 𝜑 → 𝐾 ∥ ( 𝑀 − 𝑁 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
dvds2subd.k |
⊢ ( 𝜑 → 𝐾 ∈ ℤ ) |
2 |
|
dvds2subd.m |
⊢ ( 𝜑 → 𝑀 ∈ ℤ ) |
3 |
|
dvds2subd.n |
⊢ ( 𝜑 → 𝑁 ∈ ℤ ) |
4 |
|
dvds2subd.1 |
⊢ ( 𝜑 → 𝐾 ∥ 𝑀 ) |
5 |
|
dvds2subd.2 |
⊢ ( 𝜑 → 𝐾 ∥ 𝑁 ) |
6 |
|
dvds2sub |
⊢ ( ( 𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( ( 𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁 ) → 𝐾 ∥ ( 𝑀 − 𝑁 ) ) ) |
7 |
1 2 3 6
|
syl3anc |
⊢ ( 𝜑 → ( ( 𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁 ) → 𝐾 ∥ ( 𝑀 − 𝑁 ) ) ) |
8 |
4 5 7
|
mp2and |
⊢ ( 𝜑 → 𝐾 ∥ ( 𝑀 − 𝑁 ) ) |