let X be BCI-algebra; for x, y being Element of X
for n being Nat st x \ y = x holds
(x,y) to_power n = x
let x, y be Element of X; for n being Nat st x \ y = x holds
(x,y) to_power n = x
let n be Nat; ( x \ y = x implies (x,y) to_power n = x )
defpred S1[ set ] means for m being Nat st m = $1 & m <= n holds
(x,y) to_power m = x;
assume A1:
x \ y = x
; (x,y) to_power n = x
now for k being Nat st ( for m being Nat st m = k & m <= n holds
(x,y) to_power m = x ) holds
for m being Nat st m = k + 1 & m <= n holds
(x,y) to_power (k + 1) = xlet k be
Nat;
( ( for m being Nat st m = k & m <= n holds
(x,y) to_power m = x ) implies for m being Nat st m = k + 1 & m <= n holds
(x,y) to_power (k + 1) = x )assume A2:
for
m being
Nat st
m = k &
m <= n holds
(
x,
y)
to_power m = x
;
for m being Nat st m = k + 1 & m <= n holds
(x,y) to_power (k + 1) = xlet m be
Nat;
( m = k + 1 & m <= n implies (x,y) to_power (k + 1) = x )A3:
(
x,
y)
to_power (k + 1) = ((x,y) to_power k) \ y
by Th4;
assume
(
m = k + 1 &
m <= n )
;
(x,y) to_power (k + 1) = xthen
k <= n
by NAT_1:13;
hence
(
x,
y)
to_power (k + 1) = x
by A1, A2, A3;
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
(x,y) to_power n = x
; verum