Theorem expcand

Description: Ordering relationship for exponentiation. (Contributed by Mario Carneiro, 28-May-2016)

	Ref	Expression
Hypotheses	resqcld.1	⊢ ( 𝜑 → 𝐴 ∈ ℝ )
	ltexp2d.2	⊢ ( 𝜑 → 𝑀 ∈ ℤ )
	ltexp2d.3	⊢ ( 𝜑 → 𝑁 ∈ ℤ )
	ltexp2d.4	⊢ ( 𝜑 → 1 < 𝐴 )
	expcand.5	⊢ ( 𝜑 → ( 𝐴 ↑ 𝑀 ) = ( 𝐴 ↑ 𝑁 ) )
Assertion	expcand	⊢ ( 𝜑 → 𝑀 = 𝑁 )

Proof

Step	Hyp	Ref	Expression
1		resqcld.1	⊢ ( 𝜑 → 𝐴 ∈ ℝ )
2		ltexp2d.2	⊢ ( 𝜑 → 𝑀 ∈ ℤ )
3		ltexp2d.3	⊢ ( 𝜑 → 𝑁 ∈ ℤ )
4		ltexp2d.4	⊢ ( 𝜑 → 1 < 𝐴 )
5		expcand.5	⊢ ( 𝜑 → ( 𝐴 ↑ 𝑀 ) = ( 𝐴 ↑ 𝑁 ) )
6		expcan	⊢ ( ( ( 𝐴 ∈ ℝ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ 1 < 𝐴 ) → ( ( 𝐴 ↑ 𝑀 ) = ( 𝐴 ↑ 𝑁 ) ↔ 𝑀 = 𝑁 ) )
7	1 2 3 4 6	syl31anc	⊢ ( 𝜑 → ( ( 𝐴 ↑ 𝑀 ) = ( 𝐴 ↑ 𝑁 ) ↔ 𝑀 = 𝑁 ) )
8	5 7	mpbid	⊢ ( 𝜑 → 𝑀 = 𝑁 )