Metamath Proof Explorer


Theorem dvdsexp

Description: A power divides a power with a greater exponent. (Contributed by Mario Carneiro, 23-Feb-2014)

Ref Expression
Assertion dvdsexp ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → ( 𝐴𝑀 ) ∥ ( 𝐴𝑁 ) )

Proof

Step Hyp Ref Expression
1 simp1 ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → 𝐴 ∈ ℤ )
2 uznn0sub ( 𝑁 ∈ ( ℤ𝑀 ) → ( 𝑁𝑀 ) ∈ ℕ0 )
3 2 3ad2ant3 ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → ( 𝑁𝑀 ) ∈ ℕ0 )
4 zexpcl ( ( 𝐴 ∈ ℤ ∧ ( 𝑁𝑀 ) ∈ ℕ0 ) → ( 𝐴 ↑ ( 𝑁𝑀 ) ) ∈ ℤ )
5 1 3 4 syl2anc ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → ( 𝐴 ↑ ( 𝑁𝑀 ) ) ∈ ℤ )
6 zexpcl ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0 ) → ( 𝐴𝑀 ) ∈ ℤ )
7 6 3adant3 ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → ( 𝐴𝑀 ) ∈ ℤ )
8 dvdsmul2 ( ( ( 𝐴 ↑ ( 𝑁𝑀 ) ) ∈ ℤ ∧ ( 𝐴𝑀 ) ∈ ℤ ) → ( 𝐴𝑀 ) ∥ ( ( 𝐴 ↑ ( 𝑁𝑀 ) ) · ( 𝐴𝑀 ) ) )
9 5 7 8 syl2anc ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → ( 𝐴𝑀 ) ∥ ( ( 𝐴 ↑ ( 𝑁𝑀 ) ) · ( 𝐴𝑀 ) ) )
10 1 zcnd ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → 𝐴 ∈ ℂ )
11 simp2 ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → 𝑀 ∈ ℕ0 )
12 10 11 3 expaddd ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → ( 𝐴 ↑ ( ( 𝑁𝑀 ) + 𝑀 ) ) = ( ( 𝐴 ↑ ( 𝑁𝑀 ) ) · ( 𝐴𝑀 ) ) )
13 eluzelcn ( 𝑁 ∈ ( ℤ𝑀 ) → 𝑁 ∈ ℂ )
14 13 3ad2ant3 ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → 𝑁 ∈ ℂ )
15 11 nn0cnd ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → 𝑀 ∈ ℂ )
16 14 15 npcand ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → ( ( 𝑁𝑀 ) + 𝑀 ) = 𝑁 )
17 16 oveq2d ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → ( 𝐴 ↑ ( ( 𝑁𝑀 ) + 𝑀 ) ) = ( 𝐴𝑁 ) )
18 12 17 eqtr3d ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → ( ( 𝐴 ↑ ( 𝑁𝑀 ) ) · ( 𝐴𝑀 ) ) = ( 𝐴𝑁 ) )
19 9 18 breqtrd ( ( 𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ( ℤ𝑀 ) ) → ( 𝐴𝑀 ) ∥ ( 𝐴𝑁 ) )