Theorem expcand

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

	Ref	Expression
Hypotheses	sqgt0d.1	|- ( ph -> A e. RR )
	ltexp2d.2	|- ( ph -> M e. ZZ )
	ltexp2d.3	|- ( ph -> N e. ZZ )
	ltexp2d.4	|- ( ph -> 1 < A )
	expcand.5	|- ( ph -> ( A ^ M ) = ( A ^ N ) )
Assertion	expcand	|- ( ph -> M = N )

Proof

Step	Hyp	Ref	Expression
1		sqgt0d.1	|- ( ph -> A e. RR )
2		ltexp2d.2	|- ( ph -> M e. ZZ )
3		ltexp2d.3	|- ( ph -> N e. ZZ )
4		ltexp2d.4	|- ( ph -> 1 < A )
5		expcand.5	|- ( ph -> ( A ^ M ) = ( A ^ N ) )
6		expcan	|- ( ( ( A e. RR /\ M e. ZZ /\ N e. ZZ ) /\ 1 < A ) -> ( ( A ^ M ) = ( A ^ N ) <-> M = N ) )
7	1 2 3 4 6	syl31anc	|- ( ph -> ( ( A ^ M ) = ( A ^ N ) <-> M = N ) )
8	5 7	mpbid	|- ( ph -> M = N )