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

now

(x,y to_power 0 ) \ z = x \ z by Th1;
hence (x,y to_power 0 ) \ z = (x \ z),y to_power 0 by Th1; :: thesis: verum

end;

then A1: S₁[ 0 ] ;
A2: for k being Element of NAT st S₁[k] holds
S₁[k + 1]

proof

now

let k be Element of NAT ; :: thesis: ( ( for m being Element of NAT st m = k & m <= n holds
(x,y to_power m) \ z = (x \ z),y to_power m ) implies for m being Element of NAT st m = k + 1 & m <= n holds
(x,y to_power m) \ z = (x \ z),y to_power (k + 1) )
assume A3: for m being Element of NAT st m = k & m <= n holds
(x,y to_power m) \ z = (x \ z),y to_power m ; :: thesis: for m being Element of NAT st m = k + 1 & m <= n holds
(x,y to_power m) \ z = (x \ z),y to_power (k + 1)
let m be Element of NAT ; :: thesis: ( m = k + 1 & m <= n implies (x,y to_power m) \ z = (x \ z),y to_power (k + 1) )
assume A4: ( m = k + 1 & m <= n ) ; :: thesis: (x,y to_power m) \ z = (x \ z),y to_power (k + 1)
then (x,y to_power m) \ z = ((x,y to_power k) \ y) \ z by Th4;
then A5: (x,y to_power m) \ z = ((x,y to_power k) \ z) \ y by BCIALG_1:7;
k <= n by A4, NAT_1:13;
then (x,y to_power m) \ z = ((x \ z),y to_power k) \ y by A3, A5;
hence (x,y to_power m) \ z = (x \ z),y to_power (k + 1) by Th4; :: thesis: verum

end;

hence for k being Element of NAT st S₁[k] holds
S₁[k + 1] ; :: thesis: verum

end;

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