let n be Nat; :: thesis: for X being BCI-Algebra_with_Condition(S) holds (0. X) |^ (n + 1) = 0. X
let X be BCI-Algebra_with_Condition(S); :: thesis: (0. X) |^ (n + 1) = 0. X
defpred S₁[ set ] means for m being Nat st m = $1 & m <= n holds
(0. X) |^ (m + 1) = 0. X;

now :: thesis: for k being Nat st ( for m being Nat st m = k & m <= n holds
(0. X) |^ (m + 1) = 0. X ) holds
for m being Nat st m = k + 1 & m <= n holds
(0. X) |^ (m + 1) = 0. X

let k be Nat; :: thesis: ( ( for m being Nat st m = k & m <= n holds
(0. X) |^ (m + 1) = 0. X ) implies for m being Nat st m = k + 1 & m <= n holds
(0. X) |^ (m + 1) = 0. X )
assume A1: for m being Nat st m = k & m <= n holds
(0. X) |^ (m + 1) = 0. X ; :: thesis: for m being Nat st m = k + 1 & m <= n holds
(0. X) |^ (m + 1) = 0. X
let m be Nat; :: thesis: ( m = k + 1 & m <= n implies (0. X) |^ (m + 1) = 0. X )
assume that
A2: m = k + 1 and
A3: m <= n ; :: thesis: (0. X) |^ (m + 1) = 0. X
k <= n by A2, A3, NAT_1:13;
then A4: (0. X) |^ (k + 1) = 0. X by A1;
A5: (0. X) |^ 2 = 0. X by Th24;
(0. X) |^ (m + 1) = ((0. X) |^ (k + 1)) * (0. X) by A2, Def6;
hence (0. X) |^ (m + 1) = 0. X by A5, A4, Th22; :: thesis: verum

end;

then A6: for k being Nat st S₁[k] holds
S₁[k + 1] ;
A7: S₁[ 0 ] by Th21;
for n being Nat holds S₁[n] from NAT_1:sch 2(A7, A6);
hence (0. X) |^ (n + 1) = 0. X ; :: thesis: verum