Metamath Proof Explorer


Theorem leexp2rd

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

Ref Expression
Hypotheses resqcld.1
|- ( ph -> A e. RR )
leexp2rd.2
|- ( ph -> M e. NN0 )
leexp2rd.3
|- ( ph -> N e. ( ZZ>= ` M ) )
leexp2rd.4
|- ( ph -> 0 <_ A )
leexp2rd.5
|- ( ph -> A <_ 1 )
Assertion leexp2rd
|- ( ph -> ( A ^ N ) <_ ( A ^ M ) )

Proof

Step Hyp Ref Expression
1 resqcld.1
 |-  ( ph -> A e. RR )
2 leexp2rd.2
 |-  ( ph -> M e. NN0 )
3 leexp2rd.3
 |-  ( ph -> N e. ( ZZ>= ` M ) )
4 leexp2rd.4
 |-  ( ph -> 0 <_ A )
5 leexp2rd.5
 |-  ( ph -> A <_ 1 )
6 leexp2r
 |-  ( ( ( A e. RR /\ M e. NN0 /\ N e. ( ZZ>= ` M ) ) /\ ( 0 <_ A /\ A <_ 1 ) ) -> ( A ^ N ) <_ ( A ^ M ) )
7 1 2 3 4 5 6 syl32anc
 |-  ( ph -> ( A ^ N ) <_ ( A ^ M ) )