let X be BCI-algebra; for x being Element of X
for n, m being 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; for n, m being Nat st ((0. X),x) to_power m = 0. X holds
((0. X),x) to_power (m * n) = 0. X
let n, m be Nat; ( ((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 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
; ((0. X),x) to_power (m * n) = 0. X
now for k being Nat st ( for j being Nat st j = k & j <= n holds
((0. X),x) to_power (m * j) = 0. X ) holds
for j being Nat st j = k + 1 & j <= n holds
((0. X),x) to_power (m * (k + 1)) = 0. Xlet k be
Nat;
( ( for j being Nat st j = k & j <= n holds
((0. X),x) to_power (m * j) = 0. X ) implies for j being Nat st j = k + 1 & j <= n holds
((0. X),x) to_power (m * (k + 1)) = 0. X )assume A2:
for
j being
Nat st
j = k &
j <= n holds
(
(0. X),
x)
to_power (m * j) = 0. X
;
for j being Nat st j = k + 1 & j <= n holds
((0. X),x) to_power (m * (k + 1)) = 0. Xlet j be
Nat;
( j = k + 1 & j <= n implies ((0. X),x) to_power (m * (k + 1)) = 0. X )assume
(
j = k + 1 &
j <= n )
;
((0. X),x) to_power (m * (k + 1)) = 0. Xthen 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;
verum end;
then A4:
for k being Nat st S1[k] holds
S1[k + 1]
;
A5:
S1[ 0 ]
by Th1;
for n being Nat holds S1[n]
from NAT_1:sch 2(A5, A4);
hence
((0. X),x) to_power (m * n) = 0. X
; verum