Theorem expcand

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

	Ref	Expression
Hypotheses	sqgt0d.1	$$⊢\phi \to A\in ℝ$$
	ltexp2d.2	$$⊢\phi \to M\in \mathbb{Z}$$
	ltexp2d.3	$$⊢\phi \to N\in \mathbb{Z}$$
	ltexp2d.4	$$⊢\phi \to 1<A$$
	expcand.5	$$⊢\phi \to A^M=A^N$$
Assertion	expcand	$$⊢\phi \to M=N$$

Proof

Step	Hyp	Ref	Expression
1		sqgt0d.1	$$⊢\phi \to A\in ℝ$$
2		ltexp2d.2	$$⊢\phi \to M\in \mathbb{Z}$$
3		ltexp2d.3	$$⊢\phi \to N\in \mathbb{Z}$$
4		ltexp2d.4	$$⊢\phi \to 1<A$$
5		expcand.5	$$⊢\phi \to A^M=A^N$$
6		expcan	$$⊢\left(\left(A\in ℝ\wedge M\in \mathbb{Z}\wedge N\in \mathbb{Z}\right)\wedge 1<A\right)\to \left(A^M=A^N\leftrightarrow M=N\right)$$
7	1 2 3 4 6	syl31anc	$$⊢\phi \to \left(A^M=A^N\leftrightarrow M=N\right)$$
8	5 7	mpbid	$$⊢\phi \to M=N$$