let X be BCI-algebra; for x being Element of X
for n being Element of NAT holds ((0. X),x to_power n) ` = (0. X),(x ` ) to_power n
let x be Element of X; for n being Element of NAT holds ((0. X),x to_power n) ` = (0. X),(x ` ) to_power n
let n be Element of NAT ; ((0. X),x to_power n) ` = (0. X),(x ` ) to_power n
defpred S1[ set ] means for m being Element of NAT st m = $1 & m <= n holds
((0. X),x to_power m) ` = (0. X),(x ` ) to_power m;
now let k be
Element of
NAT ;
( ( for m being Element of NAT st m = k & m <= n holds
((0. X),x to_power m) ` = (0. X),(x ` ) to_power m ) implies for m being Element of NAT st m = k + 1 & m <= n holds
((0. X),x to_power (k + 1)) ` = (0. X),(x ` ) to_power (k + 1) )assume A1:
for
m being
Element of
NAT st
m = k &
m <= n holds
((0. X),x to_power m) ` = (0. X),
(x ` ) to_power m
;
for m being Element of NAT st m = k + 1 & m <= n holds
((0. X),x to_power (k + 1)) ` = (0. X),(x ` ) to_power (k + 1)let m be
Element of
NAT ;
( m = k + 1 & m <= n implies ((0. X),x to_power (k + 1)) ` = (0. X),(x ` ) to_power (k + 1) )A2:
((0. X),x to_power (k + 1)) ` =
(((0. X),x to_power k) \ x) `
by Th4
.=
(((0. X),x to_power k) ` ) \ (x ` )
by BCIALG_1:9
;
assume
(
m = k + 1 &
m <= n )
;
((0. X),x to_power (k + 1)) ` = (0. X),(x ` ) to_power (k + 1)then
k <= n
by NAT_1:13;
hence ((0. X),x to_power (k + 1)) ` =
((0. X),(x ` ) to_power k) \ (x ` )
by A1, A2
.=
(0. X),
(x ` ) to_power (k + 1)
by Th4
;
verum end;
then A3:
for k being Element of NAT st S1[k] holds
S1[k + 1]
;
((0. X),x to_power 0 ) ` = (0. X) `
by Th1;
then
((0. X),x to_power 0 ) ` = 0. X
by BCIALG_1:2;
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) ` = (0. X),(x ` ) to_power n
; verum