let X be BCI-algebra; :: thesis: for x, z, y being Element of X
for n being Element of NAT holds (x,z to_power n) \ (y,z to_power n) <= x \ y
let x, z, y be Element of X; :: thesis: for n being Element of NAT holds (x,z to_power n) \ (y,z to_power n) <= x \ y
let n be Element of NAT ; :: thesis: (x,z to_power n) \ (y,z to_power n) <= x \ y
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) <= x \ y;
A1: 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 A2: for m being Element of NAT st m = k & m <= n holds
(x,z to_power m) \ (y,z to_power m) <= x \ y ; :: 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) <= x \ y )
assume that
A3: m = k + 1 and
A4: m <= n ; :: thesis: (x,z to_power m) \ (y,z to_power m) <= x \ y
k <= n by A3, A4, NAT_1:13;
then (x,z to_power k) \ (y,z to_power k) <= x \ y by A2;
then ((x,z to_power k) \ (y,z to_power k)) \ (x \ y) = 0. X by BCIALG_1:def 11;
then ((x,z to_power k) \ (x \ y)) \ (y,z to_power k) = 0. X by BCIALG_1:7;
then (((x,z to_power k) \ (x \ y)) \ z) \ ((y,z to_power k) \ z) = 0. X by BCIALG_1:4;
then (((x,z to_power k) \ z) \ (x \ y)) \ ((y,z to_power k) \ z) = 0. X by BCIALG_1:7;
then ((x,z to_power (k + 1)) \ (x \ y)) \ ((y,z to_power k) \ z) = 0. X by Th4;
then ((x,z to_power (k + 1)) \ (x \ y)) \ (y,z to_power (k + 1)) = 0. X by Th4;
then ((x,z to_power (k + 1)) \ (y,z to_power (k + 1))) \ (x \ y) = 0. X by BCIALG_1:7;
hence (x,z to_power m) \ (y,z to_power m) <= x \ y by A3, BCIALG_1:def 11; :: thesis: verum

end;

(x \ y) \ (x \ y) = 0. X by BCIALG_1:def 5;
then x \ y <= x \ y by BCIALG_1:def 11;
then (x,z to_power 0 ) \ y <= x \ y by Th1;
then A5: S₁[ 0 ] by Th1;
for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A5, A1);
hence (x,z to_power n) \ (y,z to_power n) <= x \ y ; :: thesis: verum