Metamath Proof Explorer


Theorem expsnnval

Description: Value of surreal exponentiation at a natural number. (Contributed by Scott Fenton, 25-Jul-2025)

Ref Expression
Assertion expsnnval Could not format assertion : No typesetting found for |- ( ( A e. No /\ N e. NN_s ) -> ( A ^su N ) = ( seq_s 1s ( x.s , ( NN_s X. { A } ) ) ` N ) ) with typecode |-

Proof

Step Hyp Ref Expression
1 nnzs N s N s
2 expsval Could not format ( ( A e. No /\ N e. ZZ_s ) -> ( A ^su N ) = if ( N = 0s , 1s , if ( 0s ( A ^su N ) = if ( N = 0s , 1s , if ( 0s
3 1 2 sylan2 Could not format ( ( A e. No /\ N e. NN_s ) -> ( A ^su N ) = if ( N = 0s , 1s , if ( 0s ( A ^su N ) = if ( N = 0s , 1s , if ( 0s
4 nnne0s N s N 0 s
5 4 neneqd N s ¬ N = 0 s
6 5 iffalsed N s if N = 0 s 1 s if 0 s < s N seq s 1 s s s × A N 1 s / su seq s 1 s s s × A + s N = if 0 s < s N seq s 1 s s s × A N 1 s / su seq s 1 s s s × A + s N
7 nnsgt0 N s 0 s < s N
8 7 iftrued N s if 0 s < s N seq s 1 s s s × A N 1 s / su seq s 1 s s s × A + s N = seq s 1 s s s × A N
9 6 8 eqtrd N s if N = 0 s 1 s if 0 s < s N seq s 1 s s s × A N 1 s / su seq s 1 s s s × A + s N = seq s 1 s s s × A N
10 9 adantl A No N s if N = 0 s 1 s if 0 s < s N seq s 1 s s s × A N 1 s / su seq s 1 s s s × A + s N = seq s 1 s s s × A N
11 3 10 eqtrd Could not format ( ( A e. No /\ N e. NN_s ) -> ( A ^su N ) = ( seq_s 1s ( x.s , ( NN_s X. { A } ) ) ` N ) ) : No typesetting found for |- ( ( A e. No /\ N e. NN_s ) -> ( A ^su N ) = ( seq_s 1s ( x.s , ( NN_s X. { A } ) ) ` N ) ) with typecode |-