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Metamath Proof Explorer

Theorem dvds1lem

Description: A lemma to assist theorems of || with one antecedent. (Contributed by Paul Chapman, 21-Mar-2011)

Ref Expression
Hypotheses dvds1lem.1 ⊢ ( 𝜑 → ( 𝐽 ∈ ℤ ∧ 𝐾 ∈ ℤ ) )
dvds1lem.2 ⊢ ( 𝜑 → ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) )
dvds1lem.3 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ ℤ ) → 𝑍 ∈ ℤ )
dvds1lem.4 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ ℤ ) → ( ( 𝑥 · 𝐽 ) = 𝐾 → ( 𝑍 · 𝑀 ) = 𝑁 ) )
Assertion dvds1lem ( 𝜑 → ( 𝐽 ∥ 𝐾 → 𝑀 ∥ 𝑁 ) )

Proof

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 ) )