Metamath Proof Explorer


Theorem ltexp2d

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

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

Proof

Step Hyp Ref Expression
1 resqcld.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 ltexp2
 |-  ( ( ( A e. RR /\ M e. ZZ /\ N e. ZZ ) /\ 1 < A ) -> ( M < N <-> ( A ^ M ) < ( A ^ N ) ) )
6 1 2 3 4 5 syl31anc
 |-  ( ph -> ( M < N <-> ( A ^ M ) < ( A ^ N ) ) )