let X be BCI-algebra; :: thesis: for x being Element of X
for n, m being 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 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 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 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 :: thesis: for k being Nat st ( for j being 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) `) ) holds
for j being 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 k be Nat; :: thesis: ( ( for j being 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 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 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 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 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 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 Nat holds S₁[n] from NAT_1:sch 2(A4, A3);
hence ((0. X),x) to_power (n + m) = (((0. X),x) to_power n) \ ((((0. X),x) to_power m) `) ; :: thesis: verum