Metamath Proof Explorer


Theorem expcand

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

Ref Expression
Hypotheses resqcld.1 φ A
ltexp2d.2 φ M
ltexp2d.3 φ N
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
2 ltexp2d.2 φ M
3 ltexp2d.3 φ N
4 ltexp2d.4 φ 1 < A
5 expcand.5 φ A M = A N
6 expcan A M N 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