let X be BCI-algebra; :: thesis: for x, y, z being Element of X

for n being 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 Nat st x <= y holds

(x,z) to_power n <= (y,z) to_power n

let n be Nat; :: thesis: ( x <= y implies (x,z) to_power n <= (y,z) to_power n )

defpred S_{1}[ set ] means for m being Nat st m = $1 & m <= n holds

(x,z) to_power m <= (y,z) to_power m;

A1: for k being Nat st S_{1}[k] holds

S_{1}[k + 1]

then (x,z) to_power 0 <= y by Th1;

then A5: S_{1}[ 0 ]
by Th1;

for n being Nat holds S_{1}[n]
from NAT_1:sch 2(A5, A1);

hence (x,z) to_power n <= (y,z) to_power n ; :: thesis: verum

for n being 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 Nat st x <= y holds

(x,z) to_power n <= (y,z) to_power n

let n be Nat; :: thesis: ( x <= y implies (x,z) to_power n <= (y,z) to_power n )

defpred S

(x,z) to_power m <= (y,z) to_power m;

A1: for k being Nat st S

S

proof

assume
x <= y
; :: thesis: (x,z) to_power n <= (y,z) to_power n
let k be Nat; :: thesis: ( S_{1}[k] implies S_{1}[k + 1] )

assume A2: for m being Nat st m = k & m <= n holds

(x,z) to_power m <= (y,z) to_power m ; :: thesis: S_{1}[k + 1]

let m be Nat; :: thesis: ( m = k + 1 & m <= n implies (x,z) to_power m <= (y,z) to_power m )

assume that

A3: m = k + 1 and

A4: m <= n ; :: thesis: (x,z) to_power m <= (y,z) to_power m

k <= n by A3, A4, NAT_1:13;

then (x,z) to_power k <= (y,z) to_power k by A2;

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 A3, Th4; :: thesis: verum

end;assume A2: for m being Nat st m = k & m <= n holds

(x,z) to_power m <= (y,z) to_power m ; :: thesis: S

let m be Nat; :: thesis: ( m = k + 1 & m <= n implies (x,z) to_power m <= (y,z) to_power m )

assume that

A3: m = k + 1 and

A4: m <= n ; :: thesis: (x,z) to_power m <= (y,z) to_power m

k <= n by A3, A4, NAT_1:13;

then (x,z) to_power k <= (y,z) to_power k by A2;

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 A3, Th4; :: thesis: verum

then (x,z) to_power 0 <= y by Th1;

then A5: S

for n being Nat holds S

hence (x,z) to_power n <= (y,z) to_power n ; :: thesis: verum