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

now

((0. X),((0. X),x to_power m) to_power 0 ) ` = (0. X) ` by Th1
.= 0. X by BCIALG_1:def 5 ;
hence ((0. X),((0. X),x to_power m) to_power 0 ) ` = (0. X),x to_power (m * 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 j being Element of NAT st j = k & j <= n holds
((0. X),((0. X),x to_power m) to_power j) ` = (0. X),x to_power (m * j) ) implies for j being Element of NAT st j = k + 1 & j <= n holds
((0. X),((0. X),x to_power m) to_power (k + 1)) ` = (0. X),x to_power (m * (k + 1)) )
assume A3: for j being Element of NAT st j = k & j <= n holds
((0. X),((0. X),x to_power m) to_power j) ` = (0. X),x to_power (m * j) ; :: thesis: for j being Element of NAT st j = k + 1 & j <= n holds
((0. X),((0. X),x to_power m) to_power (k + 1)) ` = (0. X),x to_power (m * (k + 1))
let j be Element of NAT ; :: thesis: ( j = k + 1 & j <= n implies ((0. X),((0. X),x to_power m) to_power (k + 1)) ` = (0. X),x to_power (m * (k + 1)) )
assume ( j = k + 1 & j <= n ) ; :: thesis: ((0. X),((0. X),x to_power m) to_power (k + 1)) ` = (0. X),x to_power (m * (k + 1))
then A4: k <= n by NAT_1:13;
((0. X),((0. X),x to_power m) to_power (k + 1)) ` = (((0. X),((0. X),x to_power m) to_power k) \ ((0. X),x to_power m)) ` by Th4
.= (((0. X),((0. X),x to_power m) to_power k) ` ) \ (((0. X),x to_power m) ` ) by BCIALG_1:9
.= ((0. X),x to_power (m * k)) \ (((0. X),x to_power m) ` ) by A3, A4
.= (0. X),x to_power ((m * k) + m) by Th13 ;
hence ((0. X),((0. X),x to_power m) to_power (k + 1)) ` = (0. X),x to_power (m * (k + 1)) ; :: 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),((0. X),x to_power m) to_power n) ` = (0. X),x to_power (m * n) ; :: thesis: verum