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

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 (j + m) = ((0. X),x to_power j) \ (((0. X),x to_power m) ` ) ) implies for j being Element of NAT st j = k + 1 & j <= n holds
(0. X),x to_power ((k + 1) + m) = ((0. X),x to_power (k + 1)) \ (((0. X),x to_power m) ` ) )
assume A1: for j being Element of NAT st j = k & j <= n holds
(0. X),x to_power (j + m) = ((0. X),x to_power j) \ (((0. X),x to_power m) ` ) ; :: thesis: for j being Element of NAT st j = k + 1 & j <= n holds
(0. X),x to_power ((k + 1) + m) = ((0. X),x to_power (k + 1)) \ (((0. X),x to_power m) ` )
let j be Element of NAT ; :: thesis: ( j = k + 1 & j <= n implies (0. X),x to_power ((k + 1) + m) = ((0. X),x to_power (k + 1)) \ (((0. X),x to_power m) ` ) )
assume ( j = k + 1 & j <= n ) ; :: thesis: (0. X),x to_power ((k + 1) + m) = ((0. X),x to_power (k + 1)) \ (((0. X),x to_power m) ` )
then A2: k <= n by NAT_1:13;
(0. X),x to_power ((k + m) + 1) = ((0. X),x to_power (k + m)) \ x by Th4
.= (((0. X),x to_power k) \ (((0. X),x to_power m) ` )) \ x by A1, A2
.= (((0. X),x to_power k) \ x) \ (((0. X),x to_power m) ` ) by BCIALG_1:7 ;
hence (0. X),x to_power ((k + 1) + m) = ((0. X),x to_power (k + 1)) \ (((0. X),x to_power m) ` ) by Th4; :: thesis: verum

end;

then A3: for k being Element of NAT st S₁[k] holds
S₁[k + 1] ;
(0. X),x to_power (0 + m) = (((0. X),x to_power m) ` ) ` by Th12;
then A4: S₁[ 0 ] by Th1;
for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A4, A3);
hence (0. X),x to_power (n + m) = ((0. X),x to_power n) \ (((0. X),x to_power m) ` ) ; :: thesis: verum