let X be BCI-algebra; :: thesis: for x being Element of X
for n being Element of NAT holds ((0. X),x to_power n) ` = (0. X),(x ` ) to_power n
let x be Element of X; :: thesis: for n being Element of NAT holds ((0. X),x to_power n) ` = (0. X),(x ` ) to_power n
let n be Element of NAT ; :: thesis: ((0. X),x to_power n) ` = (0. X),(x ` ) to_power n
defpred S₁[ set ] means for m being Element of NAT st m = $1 & m <= n holds
((0. X),x to_power m) ` = (0. X),(x ` ) to_power m;

now

((0. X),x to_power 0 ) ` = (0. X) ` by Th1;
then ((0. X),x to_power 0 ) ` = 0. X by BCIALG_1:2;
hence ((0. X),x to_power 0 ) ` = (0. X),(x ` ) 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
((0. X),x to_power m) ` = (0. X),(x ` ) to_power m ) implies for m being Element of NAT st m = k + 1 & m <= n holds
((0. X),x to_power (k + 1)) ` = (0. X),(x ` ) to_power (k + 1) )
assume A3: for m being Element of NAT st m = k & m <= n holds
((0. X),x to_power m) ` = (0. X),(x ` ) to_power m ; :: thesis: for m being Element of NAT st m = k + 1 & m <= n holds
((0. X),x to_power (k + 1)) ` = (0. X),(x ` ) to_power (k + 1)
let m be Element of NAT ; :: thesis: ( m = k + 1 & m <= n implies ((0. X),x to_power (k + 1)) ` = (0. X),(x ` ) to_power (k + 1) )
assume A4: ( m = k + 1 & m <= n ) ; :: thesis: ((0. X),x to_power (k + 1)) ` = (0. X),(x ` ) to_power (k + 1)
A5: ((0. X),x to_power (k + 1)) ` = (((0. X),x to_power k) \ x) ` by Th4
.= (((0. X),x to_power k) ` ) \ (x ` ) by BCIALG_1:9 ;
k <= n by A4, NAT_1:13;
hence ((0. X),x to_power (k + 1)) ` = ((0. X),(x ` ) to_power k) \ (x ` ) by A3, A5
.= (0. X),(x ` ) 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 ((0. X),x to_power n) ` = (0. X),(x ` ) to_power n ; :: thesis: verum