let X be BCI-algebra; :: thesis: for x being Element of X
for m, n being Element of NAT st ((0. X),x) to_power m = 0. X holds
((0. X),x) to_power (m * n) = 0. X
let x be Element of X; :: thesis: for m, n being Element of NAT st ((0. X),x) to_power m = 0. X holds
((0. X),x) to_power (m * n) = 0. X
let m, n be Element of NAT ; :: thesis: ( ((0. X),x) to_power m = 0. X implies ((0. X),x) to_power (m * n) = 0. X )
defpred S₁[ set ] means for j being Element of NAT st j = $1 & j <= n holds
((0. X),x) to_power (m * j) = 0. X;
assume A1: ((0. X),x) to_power m = 0. X ; :: thesis: ((0. X),x) to_power (m * n) = 0. X

now

let k be Element of NAT ; :: thesis: ( ( for j being Element of NAT st j = k & j <= n holds
((0. X),x) to_power (m * j) = 0. X ) implies for j being Element of NAT st j = k + 1 & j <= n holds
((0. X),x) to_power (m * (k + 1)) = 0. X )
assume A2: for j being Element of NAT st j = k & j <= n holds
((0. X),x) to_power (m * j) = 0. X ; :: thesis: for j being Element of NAT st j = k + 1 & j <= n holds
((0. X),x) to_power (m * (k + 1)) = 0. X
let j be Element of NAT ; :: thesis: ( j = k + 1 & j <= n implies ((0. X),x) to_power (m * (k + 1)) = 0. X )
assume ( j = k + 1 & j <= n ) ; :: thesis: ((0. X),x) to_power (m * (k + 1)) = 0. X
then A3: k <= n by NAT_1:13;
((0. X),x) to_power (m * (k + 1)) = ((0. X),x) to_power ((m * k) + m)
.= ((((0. X),x) to_power (m * k)),x) to_power m by Th10
.= ((0. X),x) to_power m by A2, A3 ;
hence ((0. X),x) to_power (m * (k + 1)) = 0. X by A1; :: thesis: verum

end;

then A4: for k being Element of NAT st S₁[k] holds
S₁[k + 1] ;
A5: S₁[ 0 ] by Th1;
for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A5, A4);
hence ((0. X),x) to_power (m * n) = 0. X ; :: thesis: verum