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

now :: thesis: for k being Nat st ( for m being Nat st m = k & m <= n holds
(x,(x \ (x \ y))) to_power m = (x,y) to_power m ) holds
for m being Nat st m = k + 1 & m <= n holds
(x,(x \ (x \ y))) to_power (k + 1) = (x,y) to_power (k + 1)

let k be Nat; :: thesis: ( ( for m being Nat st m = k & m <= n holds
(x,(x \ (x \ y))) to_power m = (x,y) to_power m ) implies for m being Nat st m = k + 1 & m <= n holds
(x,(x \ (x \ y))) to_power (k + 1) = (x,y) to_power (k + 1) )
assume A1: for m being Nat st m = k & m <= n holds
(x,(x \ (x \ y))) to_power m = (x,y) to_power m ; :: thesis: for m being Nat st m = k + 1 & m <= n holds
(x,(x \ (x \ y))) to_power (k + 1) = (x,y) to_power (k + 1)
let m be Nat; :: thesis: ( m = k + 1 & m <= n implies (x,(x \ (x \ y))) to_power (k + 1) = (x,y) to_power (k + 1) )
A2: (x,(x \ (x \ y))) to_power (k + 1) = ((x,(x \ (x \ y))) to_power k) \ (x \ (x \ y)) by Th4;
assume ( m = k + 1 & m <= n ) ; :: thesis: (x,(x \ (x \ y))) to_power (k + 1) = (x,y) to_power (k + 1)
then k <= n by NAT_1:13;
hence (x,(x \ (x \ y))) to_power (k + 1) = ((x,y) to_power k) \ (x \ (x \ y)) by A1, A2
.= ((x \ (x \ (x \ y))),y) to_power k by Th7
.= ((x \ y),y) to_power k by BCIALG_1:8
.= ((x,y) to_power k) \ y by Th7
.= (x,y) to_power (k + 1) by Th4 ;
:: thesis: verum

end;

then A3: for k being Nat st S₁[k] holds
S₁[k + 1] ;
(x,(x \ (x \ y))) to_power 0 = x by Th1;
then A4: S₁[ 0 ] by Th1;
for n being Nat holds S₁[n] from NAT_1:sch 2(A4, A3);
hence (x,(x \ (x \ y))) to_power n = (x,y) to_power n ; :: thesis: verum