Step |
Hyp |
Ref |
Expression |
1 |
|
dvds1lem.1 |
⊢ ( 𝜑 → ( 𝐽 ∈ ℤ ∧ 𝐾 ∈ ℤ ) ) |
2 |
|
dvds1lem.2 |
⊢ ( 𝜑 → ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ) |
3 |
|
dvds1lem.3 |
⊢ ( ( 𝜑 ∧ 𝑥 ∈ ℤ ) → 𝑍 ∈ ℤ ) |
4 |
|
dvds1lem.4 |
⊢ ( ( 𝜑 ∧ 𝑥 ∈ ℤ ) → ( ( 𝑥 · 𝐽 ) = 𝐾 → ( 𝑍 · 𝑀 ) = 𝑁 ) ) |
5 |
|
oveq1 |
⊢ ( 𝑧 = 𝑍 → ( 𝑧 · 𝑀 ) = ( 𝑍 · 𝑀 ) ) |
6 |
5
|
eqeq1d |
⊢ ( 𝑧 = 𝑍 → ( ( 𝑧 · 𝑀 ) = 𝑁 ↔ ( 𝑍 · 𝑀 ) = 𝑁 ) ) |
7 |
6
|
rspcev |
⊢ ( ( 𝑍 ∈ ℤ ∧ ( 𝑍 · 𝑀 ) = 𝑁 ) → ∃ 𝑧 ∈ ℤ ( 𝑧 · 𝑀 ) = 𝑁 ) |
8 |
3 4 7
|
syl6an |
⊢ ( ( 𝜑 ∧ 𝑥 ∈ ℤ ) → ( ( 𝑥 · 𝐽 ) = 𝐾 → ∃ 𝑧 ∈ ℤ ( 𝑧 · 𝑀 ) = 𝑁 ) ) |
9 |
8
|
rexlimdva |
⊢ ( 𝜑 → ( ∃ 𝑥 ∈ ℤ ( 𝑥 · 𝐽 ) = 𝐾 → ∃ 𝑧 ∈ ℤ ( 𝑧 · 𝑀 ) = 𝑁 ) ) |
10 |
|
divides |
⊢ ( ( 𝐽 ∈ ℤ ∧ 𝐾 ∈ ℤ ) → ( 𝐽 ∥ 𝐾 ↔ ∃ 𝑥 ∈ ℤ ( 𝑥 · 𝐽 ) = 𝐾 ) ) |
11 |
1 10
|
syl |
⊢ ( 𝜑 → ( 𝐽 ∥ 𝐾 ↔ ∃ 𝑥 ∈ ℤ ( 𝑥 · 𝐽 ) = 𝐾 ) ) |
12 |
|
divides |
⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 ∥ 𝑁 ↔ ∃ 𝑧 ∈ ℤ ( 𝑧 · 𝑀 ) = 𝑁 ) ) |
13 |
2 12
|
syl |
⊢ ( 𝜑 → ( 𝑀 ∥ 𝑁 ↔ ∃ 𝑧 ∈ ℤ ( 𝑧 · 𝑀 ) = 𝑁 ) ) |
14 |
9 11 13
|
3imtr4d |
⊢ ( 𝜑 → ( 𝐽 ∥ 𝐾 → 𝑀 ∥ 𝑁 ) ) |