Metamath Proof Explorer


Theorem expsubd

Description: Exponent subtraction law for nonnegative integer exponentiation. (Contributed by Mario Carneiro, 28-May-2016)

Ref Expression
Hypotheses expcld.1 ( 𝜑𝐴 ∈ ℂ )
sqrecd.1 ( 𝜑𝐴 ≠ 0 )
expclzd.3 ( 𝜑𝑁 ∈ ℤ )
expsubd.3 ( 𝜑𝑀 ∈ ℤ )
Assertion expsubd ( 𝜑 → ( 𝐴 ↑ ( 𝑀𝑁 ) ) = ( ( 𝐴𝑀 ) / ( 𝐴𝑁 ) ) )

Proof

Step Hyp Ref Expression
1 expcld.1 ( 𝜑𝐴 ∈ ℂ )
2 sqrecd.1 ( 𝜑𝐴 ≠ 0 )
3 expclzd.3 ( 𝜑𝑁 ∈ ℤ )
4 expsubd.3 ( 𝜑𝑀 ∈ ℤ )
5 expsub ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ) → ( 𝐴 ↑ ( 𝑀𝑁 ) ) = ( ( 𝐴𝑀 ) / ( 𝐴𝑁 ) ) )
6 1 2 4 3 5 syl22anc ( 𝜑 → ( 𝐴 ↑ ( 𝑀𝑁 ) ) = ( ( 𝐴𝑀 ) / ( 𝐴𝑁 ) ) )