Metamath Proof Explorer

Theorem nndivtr

Description: Transitive property of divisibility: if A divides B and B divides C , then A divides C . Typically, C would be an integer, although the theorem holds for complex C . (Contributed by NM, 3-May-2005)

Ref Expression
Assertion nndivtr ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℂ ) ∧ ( ( 𝐵 / 𝐴 ) ∈ ℕ ∧ ( 𝐶 / 𝐵 ) ∈ ℕ ) ) → ( 𝐶 / 𝐴 ) ∈ ℕ )

Proof

Step Hyp Ref Expression
1 nnmulcl ⊢ ( ( ( 𝐵 / 𝐴 ) ∈ ℕ ∧ ( 𝐶 / 𝐵 ) ∈ ℕ ) → ( ( 𝐵 / 𝐴 ) · ( 𝐶 / 𝐵 ) ) ∈ ℕ )
2 nncn ⊢ ( 𝐵 ∈ ℕ → 𝐵 ∈ ℂ )
3 2 3ad2ant2 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℂ ) → 𝐵 ∈ ℂ )
4 simp3 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℂ ) → 𝐶 ∈ ℂ )
5 nncn ⊢ ( 𝐴 ∈ ℕ → 𝐴 ∈ ℂ )
6 nnne0 ⊢ ( 𝐴 ∈ ℕ → 𝐴 ≠ 0 )
7 5 6 jca ⊢ ( 𝐴 ∈ ℕ → ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) )
8 7 3ad2ant1 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) )
9 nnne0 ⊢ ( 𝐵 ∈ ℕ → 𝐵 ≠ 0 )
10 2 9 jca ⊢ ( 𝐵 ∈ ℕ → ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) )
11 10 3ad2ant2 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) )
12 divmul24 ⊢ ( ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) ) → ( ( 𝐵 / 𝐴 ) · ( 𝐶 / 𝐵 ) ) = ( ( 𝐵 / 𝐵 ) · ( 𝐶 / 𝐴 ) ) )
13 3 4 8 11 12 syl22anc ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 / 𝐴 ) · ( 𝐶 / 𝐵 ) ) = ( ( 𝐵 / 𝐵 ) · ( 𝐶 / 𝐴 ) ) )
14 2 9 dividd ⊢ ( 𝐵 ∈ ℕ → ( 𝐵 / 𝐵 ) = 1 )
15 14 oveq1d ⊢ ( 𝐵 ∈ ℕ → ( ( 𝐵 / 𝐵 ) · ( 𝐶 / 𝐴 ) ) = ( 1 · ( 𝐶 / 𝐴 ) ) )
16 15 3ad2ant2 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 / 𝐵 ) · ( 𝐶 / 𝐴 ) ) = ( 1 · ( 𝐶 / 𝐴 ) ) )
17 divcl ⊢ ( ( 𝐶 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 𝐶 / 𝐴 ) ∈ ℂ )
18 17 3expb ⊢ ( ( 𝐶 ∈ ℂ ∧ ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ) → ( 𝐶 / 𝐴 ) ∈ ℂ )
19 7 18 sylan2 ⊢ ( ( 𝐶 ∈ ℂ ∧ 𝐴 ∈ ℕ ) → ( 𝐶 / 𝐴 ) ∈ ℂ )
20 19 ancoms ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐶 ∈ ℂ ) → ( 𝐶 / 𝐴 ) ∈ ℂ )
21 20 mullidd ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐶 ∈ ℂ ) → ( 1 · ( 𝐶 / 𝐴 ) ) = ( 𝐶 / 𝐴 ) )
22 21 3adant2 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℂ ) → ( 1 · ( 𝐶 / 𝐴 ) ) = ( 𝐶 / 𝐴 ) )
23 13 16 22 3eqtrd ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 / 𝐴 ) · ( 𝐶 / 𝐵 ) ) = ( 𝐶 / 𝐴 ) )
24 23 eleq1d ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐵 / 𝐴 ) · ( 𝐶 / 𝐵 ) ) ∈ ℕ ↔ ( 𝐶 / 𝐴 ) ∈ ℕ ) )
25 1 24 imbitrid ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐵 / 𝐴 ) ∈ ℕ ∧ ( 𝐶 / 𝐵 ) ∈ ℕ ) → ( 𝐶 / 𝐴 ) ∈ ℕ ) )
26 25 imp ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℂ ) ∧ ( ( 𝐵 / 𝐴 ) ∈ ℕ ∧ ( 𝐶 / 𝐵 ) ∈ ℕ ) ) → ( 𝐶 / 𝐴 ) ∈ ℕ )