let X be BCI-algebra; :: thesis: for x, y, z being Element of X
for n being Element of NAT st x <= y holds
x,z to_power n <= y,z to_power n
let x, y, z be Element of X; :: thesis: for n being Element of NAT st x <= y holds
x,z to_power n <= y,z to_power n
let n be Element of NAT ; :: thesis: ( x <= y implies x,z to_power n <= y,z to_power n )
assume A1: x <= y ; :: thesis: x,z to_power n <= y,z to_power n
defpred S₁[ set ] means for m being Element of NAT st m = $1 & m <= n holds
x,z to_power m <= y,z to_power m;

now

x,z to_power 0 <= y by A1, Th1;
hence x,z to_power 0 <= y,z to_power 0 by Th1; :: thesis: verum

end;

then A2: S₁[ 0 ] ;
A3: for k being Element of NAT st S₁[k] holds
S₁[k + 1]

proof

let k be Element of NAT ; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A4: for m being Element of NAT st m = k & m <= n holds
x,z to_power m <= y,z to_power m ; :: thesis: S₁[k + 1]
let m be Element of NAT ; :: thesis: ( m = k + 1 & m <= n implies x,z to_power m <= y,z to_power m )
assume A5: ( m = k + 1 & m <= n ) ; :: thesis: x,z to_power m <= y,z to_power m
then k <= n by NAT_1:13;
then x,z to_power k <= y,z to_power k by A4;
then (x,z to_power k) \ z <= (y,z to_power k) \ z by BCIALG_1:5;
then x,z to_power (k + 1) <= (y,z to_power k) \ z by Th4;
hence x,z to_power m <= y,z to_power m by A5, Th4; :: thesis: verum

end;

for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A2, A3);
hence x,z to_power n <= y,z to_power n ; :: thesis: verum