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Metamath Proof Explorer

Theorem dvdslcm

Description: The lcm of two integers is divisible by each of them. (Contributed by Steve Rodriguez, 20-Jan-2020)

Ref Expression
Assertion dvdslcm ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 ∥ ( 𝑀 lcm 𝑁 ) ∧ 𝑁 ∥ ( 𝑀 lcm 𝑁 ) ) )

Proof

Step Hyp Ref Expression
1 dvds0 ⊢ ( 𝑀 ∈ ℤ → 𝑀 ∥ 0 )
2 1 ad2antrr ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ( 𝑀 = 0 ∨ 𝑁 = 0 ) ) → 𝑀 ∥ 0 )
3 oveq1 ⊢ ( 𝑀 = 0 → ( 𝑀 lcm 𝑁 ) = ( 0 lcm 𝑁 ) )
4 0z ⊢ 0 ∈ ℤ
5 lcmcom ⊢ ( ( 𝑁 ∈ ℤ ∧ 0 ∈ ℤ ) → ( 𝑁 lcm 0 ) = ( 0 lcm 𝑁 ) )
6 4 5 mpan2 ⊢ ( 𝑁 ∈ ℤ → ( 𝑁 lcm 0 ) = ( 0 lcm 𝑁 ) )
7 lcm0val ⊢ ( 𝑁 ∈ ℤ → ( 𝑁 lcm 0 ) = 0 )
8 6 7 eqtr3d ⊢ ( 𝑁 ∈ ℤ → ( 0 lcm 𝑁 ) = 0 )
9 3 8 sylan9eqr ⊢ ( ( 𝑁 ∈ ℤ ∧ 𝑀 = 0 ) → ( 𝑀 lcm 𝑁 ) = 0 )
10 9 adantll ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ 𝑀 = 0 ) → ( 𝑀 lcm 𝑁 ) = 0 )
11 oveq2 ⊢ ( 𝑁 = 0 → ( 𝑀 lcm 𝑁 ) = ( 𝑀 lcm 0 ) )
12 lcm0val ⊢ ( 𝑀 ∈ ℤ → ( 𝑀 lcm 0 ) = 0 )
13 11 12 sylan9eqr ⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 = 0 ) → ( 𝑀 lcm 𝑁 ) = 0 )
14 13 adantlr ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ 𝑁 = 0 ) → ( 𝑀 lcm 𝑁 ) = 0 )
15 10 14 jaodan ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ( 𝑀 = 0 ∨ 𝑁 = 0 ) ) → ( 𝑀 lcm 𝑁 ) = 0 )
16 2 15 breqtrrd ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ( 𝑀 = 0 ∨ 𝑁 = 0 ) ) → 𝑀 ∥ ( 𝑀 lcm 𝑁 ) )
17 dvds0 ⊢ ( 𝑁 ∈ ℤ → 𝑁 ∥ 0 )
18 17 ad2antlr ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ( 𝑀 = 0 ∨ 𝑁 = 0 ) ) → 𝑁 ∥ 0 )
19 18 15 breqtrrd ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ( 𝑀 = 0 ∨ 𝑁 = 0 ) ) → 𝑁 ∥ ( 𝑀 lcm 𝑁 ) )
20 16 19 jca ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ( 𝑀 = 0 ∨ 𝑁 = 0 ) ) → ( 𝑀 ∥ ( 𝑀 lcm 𝑁 ) ∧ 𝑁 ∥ ( 𝑀 lcm 𝑁 ) ) )
21 lcmcllem ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ¬ ( 𝑀 = 0 ∨ 𝑁 = 0 ) ) → ( 𝑀 lcm 𝑁 ) ∈ { 𝑛 ∈ ℕ ∣ ( 𝑀 ∥ 𝑛 ∧ 𝑁 ∥ 𝑛 ) } )
22 lcmn0cl ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ¬ ( 𝑀 = 0 ∨ 𝑁 = 0 ) ) → ( 𝑀 lcm 𝑁 ) ∈ ℕ )
23 breq2 ⊢ ( 𝑛 = ( 𝑀 lcm 𝑁 ) → ( 𝑀 ∥ 𝑛 ↔ 𝑀 ∥ ( 𝑀 lcm 𝑁 ) ) )
24 breq2 ⊢ ( 𝑛 = ( 𝑀 lcm 𝑁 ) → ( 𝑁 ∥ 𝑛 ↔ 𝑁 ∥ ( 𝑀 lcm 𝑁 ) ) )
25 23 24 anbi12d ⊢ ( 𝑛 = ( 𝑀 lcm 𝑁 ) → ( ( 𝑀 ∥ 𝑛 ∧ 𝑁 ∥ 𝑛 ) ↔ ( 𝑀 ∥ ( 𝑀 lcm 𝑁 ) ∧ 𝑁 ∥ ( 𝑀 lcm 𝑁 ) ) ) )
26 25 elrab3 ⊢ ( ( 𝑀 lcm 𝑁 ) ∈ ℕ → ( ( 𝑀 lcm 𝑁 ) ∈ { 𝑛 ∈ ℕ ∣ ( 𝑀 ∥ 𝑛 ∧ 𝑁 ∥ 𝑛 ) } ↔ ( 𝑀 ∥ ( 𝑀 lcm 𝑁 ) ∧ 𝑁 ∥ ( 𝑀 lcm 𝑁 ) ) ) )
27 22 26 syl ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ¬ ( 𝑀 = 0 ∨ 𝑁 = 0 ) ) → ( ( 𝑀 lcm 𝑁 ) ∈ { 𝑛 ∈ ℕ ∣ ( 𝑀 ∥ 𝑛 ∧ 𝑁 ∥ 𝑛 ) } ↔ ( 𝑀 ∥ ( 𝑀 lcm 𝑁 ) ∧ 𝑁 ∥ ( 𝑀 lcm 𝑁 ) ) ) )
28 21 27 mpbid ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ¬ ( 𝑀 = 0 ∨ 𝑁 = 0 ) ) → ( 𝑀 ∥ ( 𝑀 lcm 𝑁 ) ∧ 𝑁 ∥ ( 𝑀 lcm 𝑁 ) ) )
29 20 28 pm2.61dan ⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 ∥ ( 𝑀 lcm 𝑁 ) ∧ 𝑁 ∥ ( 𝑀 lcm 𝑁 ) ) )