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

now

z \ z = 0. X by BCIALG_1:def 5;
then z <= z by BCIALG_1:def 11;
then z,y to_power 0 <= z by Th1;
hence z,y to_power 0 <= z,x 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
z,y to_power m <= z,x to_power m ; :: thesis: S₁[k + 1]
let m be Element of NAT ; :: thesis: ( m = k + 1 & m <= n implies z,y to_power m <= z,x to_power m )
assume A5: ( m = k + 1 & m <= n ) ; :: thesis: z,y to_power m <= z,x to_power m
then k <= n by NAT_1:13;
then z,y to_power k <= z,x to_power k by A4;
then (z,y to_power k) \ y <= (z,x to_power k) \ y by BCIALG_1:5;
then z,y to_power (k + 1) <= (z,x to_power k) \ y by Th4;
then A6: (z,y to_power (k + 1)) \ ((z,x to_power k) \ y) = 0. X by BCIALG_1:def 11;
(z,x to_power k) \ y <= (z,x to_power k) \ x by A1, BCIALG_1:5;
then (z,x to_power k) \ y <= z,x to_power (k + 1) by Th4;
then ((z,x to_power k) \ y) \ (z,x to_power (k + 1)) = 0. X by BCIALG_1:def 11;
then (z,y to_power (k + 1)) \ (z,x to_power (k + 1)) = 0. X by A6, BCIALG_1:3;
hence z,y to_power m <= z,x to_power m by A5, BCIALG_1:def 11; :: thesis: verum

end;

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