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 )
assume A1: (0. X),x to_power m = 0. X ; :: thesis: (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;
A2: S₁[ 0 ] by Th1;
A3: 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),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 A4: 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 A5: 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 A4, A5 ;
hence (0. X),x to_power (m * (k + 1)) = 0. X by A1; :: 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(A2, A3);
hence (0. X),x to_power (m * n) = 0. X ; :: thesis: verum