let X be BCI-algebra; for x, y being Element of X
for n, m being Element of NAT holds (x,y to_power n),y to_power m = x,y to_power (n + m)
let x, y be Element of X; for n, m being Element of NAT holds (x,y to_power n),y to_power m = x,y to_power (n + m)
let n, m be Element of NAT ; (x,y to_power n),y to_power m = x,y to_power (n + m)
defpred S1[ set ] means for m1 being Element of NAT st m1 = $1 & m1 <= n holds
(x,y to_power m1),y to_power m = x,y to_power (m1 + m);
now let k be
Element of
NAT ;
( ( for m1 being Element of NAT st m1 = k & m1 <= n holds
(x,y to_power m1),y to_power m = x,y to_power (m1 + m) ) implies for m1 being Element of NAT st m1 = k + 1 & m1 <= n holds
(x,y to_power m1),y to_power m = x,y to_power (m1 + m) )assume A1:
for
m1 being
Element of
NAT st
m1 = k &
m1 <= n holds
(x,y to_power m1),
y to_power m = x,
y to_power (m1 + m)
;
for m1 being Element of NAT st m1 = k + 1 & m1 <= n holds
(x,y to_power m1),y to_power m = x,y to_power (m1 + m)let m1 be
Element of
NAT ;
( m1 = k + 1 & m1 <= n implies (x,y to_power m1),y to_power m = x,y to_power (m1 + m) )assume that A2:
m1 = k + 1
and A3:
m1 <= n
;
(x,y to_power m1),y to_power m = x,y to_power (m1 + m)
k <= n
by A2, A3, NAT_1:13;
then
(x,y to_power k),
y to_power m = x,
y to_power (k + m)
by A1;
then
((x,y to_power k),y to_power m) \ y = x,
y to_power ((k + m) + 1)
by Th4;
then
((x,y to_power k) \ y),
y to_power m = x,
y to_power ((k + m) + 1)
by Th7;
hence
(x,y to_power m1),
y to_power m = x,
y to_power (m1 + m)
by A2, Th4;
verum end;
then A4:
for k being Element of NAT st S1[k] holds
S1[k + 1]
;
A5:
S1[ 0 ]
by Th1;
for n being Element of NAT holds S1[n]
from NAT_1:sch 1(A5, A4);
hence
(x,y to_power n),y to_power m = x,y to_power (n + m)
; verum