let i, j, m, n be Element of NAT ; for X being BCK-algebra of i,j,m,n st n = 0 & i <> 0 holds
X is BCK-algebra of 0 , 0 , 0 , 0
let X be BCK-algebra of i,j,m,n; ( n = 0 & i <> 0 implies X is BCK-algebra of 0 , 0 , 0 , 0 )
assume that
A1:
n = 0
and
A2:
i <> 0
; X is BCK-algebra of 0 , 0 , 0 , 0
for x, y being Element of X holds Polynom 0 ,j,x,y = Polynom 0 ,0 ,y,x
proof
let x,
y be
Element of
X;
Polynom 0 ,j,x,y = Polynom 0 ,0 ,y,x
A3:
Polynom i,
j,
x,
y = Polynom m,
n,
y,
x
by Def3;
A4:
m + 1
>= n + 1
by A1, XREAL_1:8;
(
y,
(y \ x) to_power (n + 1) = y,
(y \ x) to_power (i + 1) &
n + 1
< i + 1 )
by A1, A2, Th19, XREAL_1:8;
then
y,
(y \ x) to_power (m + 1) = y,
(y \ x) to_power (0 + 1)
by A1, A4, Th6;
hence
Polynom 0 ,
j,
x,
y = Polynom 0 ,
0 ,
y,
x
by A1, A3, Th19;
verum
end;
then reconsider X = X as BCK-algebra of 0 ,j, 0 , 0 by Def3;
A5:
for x, y being Element of X holds Polynom 0 ,0 ,y,x <= Polynom 0 ,0 ,x,y
proof
let x,
y be
Element of
X;
Polynom 0 ,0 ,y,x <= Polynom 0 ,0 ,x,y
Polynom 0 ,
j,
x,
y = Polynom 0 ,
0 ,
y,
x
by Def3;
hence
Polynom 0 ,
0 ,
y,
x <= Polynom 0 ,
0 ,
x,
y
by Th5;
verum
end;
for x, y being Element of X holds Polynom 0 ,0 ,y,x = Polynom 0 ,0 ,x,y
proof
let x,
y be
Element of
X;
Polynom 0 ,0 ,y,x = Polynom 0 ,0 ,x,y
Polynom 0 ,
0 ,
x,
y <= Polynom 0 ,
0 ,
y,
x
by A5;
then A6:
(Polynom 0 ,0 ,x,y) \ (Polynom 0 ,0 ,y,x) = 0. X
by BCIALG_1:def 11;
Polynom 0 ,
0 ,
y,
x <= Polynom 0 ,
0 ,
x,
y
by A5;
then
(Polynom 0 ,0 ,y,x) \ (Polynom 0 ,0 ,x,y) = 0. X
by BCIALG_1:def 11;
hence
Polynom 0 ,
0 ,
y,
x = Polynom 0 ,
0 ,
x,
y
by A6, BCIALG_1:def 7;
verum
end;
hence
X is BCK-algebra of 0 , 0 , 0 , 0
by Def3; verum