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

now

x \ x = 0. X by BCIALG_1:def 5;
then x <= x by BCIALG_1:def 11;
then x,(y \ x) to_power 0 <= x by Th1;
hence (x,(x \ y) to_power 0 ),(y \ x) to_power 0 <= x by Th1; :: thesis: verum

end;

then A1: S₁[ 0 ] ;
A2: 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 A3: for m being Element of NAT st m = k & m <= n holds
(x,(x \ y) to_power m),(y \ x) to_power m <= x ; :: thesis: S₁[k + 1]
let m be Element of NAT ; :: thesis: ( m = k + 1 & m <= n implies (x,(x \ y) to_power m),(y \ x) to_power m <= x )
assume A4: ( m = k + 1 & m <= n ) ; :: thesis: (x,(x \ y) to_power m),(y \ x) to_power m <= x
then k <= n by NAT_1:13;
then (x,(x \ y) to_power k),(y \ x) to_power k <= x by A3;
then ((x,(x \ y) to_power k),(y \ x) to_power k) \ x = 0. X by BCIALG_1:def 11;
then (((x,(x \ y) to_power k) \ x),(y \ x) to_power k) \ (y \ x) = (y \ x) ` by Th7;
then (((x,(x \ y) to_power k) \ x),(y \ x) to_power (k + 1)) \ (x \ y) = ((y \ x) ` ) \ (x \ y) by Th4;
then (((x,(x \ y) to_power k) \ x) \ (x \ y)),(y \ x) to_power (k + 1) = ((y \ x) ` ) \ (x \ y) by Th7;
then (((x,(x \ y) to_power k) \ (x \ y)) \ x),(y \ x) to_power (k + 1) = ((y \ x) ` ) \ (x \ y) by BCIALG_1:7;
then ((x,(x \ y) to_power (k + 1)) \ x),(y \ x) to_power (k + 1) = ((y \ x) ` ) \ (x \ y) by Th4;
then ((x,(x \ y) to_power (k + 1)) \ x),(y \ x) to_power (k + 1) = ((y \ y) \ (y \ x)) \ (x \ y) by BCIALG_1:def 5;
then ((x,(x \ y) to_power (k + 1)) \ x),(y \ x) to_power (k + 1) = 0. X by BCIALG_1:1;
then ((x,(x \ y) to_power (k + 1)),(y \ x) to_power (k + 1)) \ x = 0. X by Th7;
hence (x,(x \ y) to_power m),(y \ x) to_power m <= x by A4, BCIALG_1:def 11; :: thesis: verum

end;

for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A1, A2);
hence (x,(x \ y) to_power n),(y \ x) to_power n <= x ; :: thesis: verum