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Metamath Proof Explorer

Theorem expsubd

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

Ref Expression
Hypotheses expcld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
sqrecd.1 ⊢ ( 𝜑 → 𝐴 ≠ 0 )
expclzd.3 ⊢ ( 𝜑 → 𝑁 ∈ ℤ )
expsubd.3 ⊢ ( 𝜑 → 𝑀 ∈ ℤ )
Assertion expsubd ( 𝜑 → ( 𝐴 ↑ ( 𝑀 − 𝑁 ) ) = ( ( 𝐴 ↑ 𝑀 ) / ( 𝐴 ↑ 𝑁 ) ) )

Proof

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