let X be BCI-algebra; :: thesis: for x, z, y being Element of X
for n being Element of NAT holds (x,z to_power n) \ (y,z to_power n) <= x \ y
let x, z, y be Element of X; :: thesis: for n being Element of NAT holds (x,z to_power n) \ (y,z to_power n) <= x \ y
let n be Element of NAT ; :: thesis: (x,z to_power n) \ (y,z to_power n) <= x \ y
defpred S1[ set ] means for m being Element of NAT st m = $1 & m <= n holds
(x,z to_power m) \ (y,z to_power m) <= x \ y;
then A1:
S1[ 0 ]
;
A2:
for k being Element of NAT st S1[k] holds
S1[k + 1]
proof
let k be
Element of
NAT ;
:: thesis: ( S1[k] implies S1[k + 1] )
assume A3:
for
m being
Element of
NAT st
m = k &
m <= n holds
(x,z to_power m) \ (y,z to_power m) <= x \ y
;
:: thesis: S1[k + 1]
let m be
Element of
NAT ;
:: thesis: ( m = k + 1 & m <= n implies (x,z to_power m) \ (y,z to_power m) <= x \ y )
assume A4:
(
m = k + 1 &
m <= n )
;
:: thesis: (x,z to_power m) \ (y,z to_power m) <= x \ y
then
k <= n
by NAT_1:13;
then
(x,z to_power k) \ (y,z to_power k) <= x \ y
by A3;
then
((x,z to_power k) \ (y,z to_power k)) \ (x \ y) = 0. X
by BCIALG_1:def 11;
then
((x,z to_power k) \ (x \ y)) \ (y,z to_power k) = 0. X
by BCIALG_1:7;
then
(((x,z to_power k) \ (x \ y)) \ z) \ ((y,z to_power k) \ z) = 0. X
by BCIALG_1:4;
then
(((x,z to_power k) \ z) \ (x \ y)) \ ((y,z to_power k) \ z) = 0. X
by BCIALG_1:7;
then
((x,z to_power (k + 1)) \ (x \ y)) \ ((y,z to_power k) \ z) = 0. X
by Th4;
then
((x,z to_power (k + 1)) \ (x \ y)) \ (y,z to_power (k + 1)) = 0. X
by Th4;
then
((x,z to_power (k + 1)) \ (y,z to_power (k + 1))) \ (x \ y) = 0. X
by BCIALG_1:7;
hence
(x,z to_power m) \ (y,z to_power m) <= x \ y
by A4, BCIALG_1:def 11;
:: thesis: verum
end;
for n being Element of NAT holds S1[n]
from NAT_1:sch 1(A1, A2);
hence
(x,z to_power n) \ (y,z to_power n) <= x \ y
; :: thesis: verum