let i, j, m, n be Nat; for X being BCK-algebra of i,j,m,n st m = 0 & j > 0 holds
X is BCK-algebra of 0 ,1, 0 ,1
let X be BCK-algebra of i,j,m,n; ( m = 0 & j > 0 implies X is BCK-algebra of 0 ,1, 0 ,1 )
reconsider X = X as BCK-algebra of i + 1,j,m,n + 1 by Th17;
assume that
A1:
m = 0
and
A2:
j > 0
; X is BCK-algebra of 0 ,1, 0 ,1
for x, y being Element of X holds Polynom (0,1,x,y) = Polynom (0,1,y,x)
proof
let x,
y be
Element of
X;
Polynom (0,1,x,y) = Polynom (0,1,y,x)
A3:
(i + 1) + 1
> (m + 1) + 0
by A1, XREAL_1:8;
A4:
(
((x,(x \ y)) to_power (0 + 1)),
(y \ x))
to_power (j + 1) = (
((x,(x \ y)) to_power (0 + 1)),
(y \ x))
to_power (m + 1)
by Th20;
A5:
j + 1
> m + 1
by A1, A2, XREAL_1:6;
(
n + 1
>= m + 1 & (
((y,(y \ x)) to_power (0 + 1)),
(x \ y))
to_power (j + 1) = (
((y,(y \ x)) to_power (0 + 1)),
(x \ y))
to_power (m + 1) )
by A1, Th20, XREAL_1:6;
then A6:
(
((y,(y \ x)) to_power (0 + 1)),
(x \ y))
to_power (0 + 1) = (
((y,(y \ x)) to_power (0 + 1)),
(x \ y))
to_power (n + 1)
by A1, A5, Th6;
(
Polynom (
(i + 1),
j,
x,
y)
= Polynom (
m,
(n + 1),
y,
x) & (
x,
(x \ y))
to_power (j + 1) = (
x,
(x \ y))
to_power (m + 1) )
by Def3, Th20;
then
(
((x,(x \ y)) to_power (0 + 1)),
(y \ x))
to_power j = (
((y,(y \ x)) to_power (0 + 1)),
(x \ y))
to_power (n + 1)
by A1, A5, A3, Th6;
hence
Polynom (
0,1,
x,
y)
= Polynom (
0,1,
y,
x)
by A1, A5, A6, A4, Th6, NAT_1:14;
verum
end;
hence
X is BCK-algebra of 0 ,1, 0 ,1
by Def3; verum