set I = I_I3 ;
ex y being Element of [.0,1.] st I_I3 . (1,y) <> y
then T1:
not I_I3 is satisfying_(NP)
by FUZIMPL2:def 1;
for x, y, z being Element of [.0,1.] holds I_I3 . (x,(I_I3 . (y,z))) = I_I3 . (y,(I_I3 . (x,z)))
proof
let x,
y,
z be
Element of
[.0,1.];
I_I3 . (x,(I_I3 . (y,z))) = I_I3 . (y,(I_I3 . (x,z)))
per cases
( z <> 0 or ( z = 0 & y <> 0 & x <> 0 ) or ( z = 0 & y <> 0 & x = 0 ) or ( z = 0 & y = 0 & x = 0 ) or ( z = 0 & y = 0 & x <> 0 ) )
;
suppose
z <> 0
;
I_I3 . (x,(I_I3 . (y,z))) = I_I3 . (y,(I_I3 . (x,z)))then D2:
(
I_I3 . (
y,
z)
= 1 &
I_I3 . (
x,
z)
= 1 )
by II3Def;
then I_I3 . (
x,
(I_I3 . (y,z))) =
1
by II3Def
.=
I_I3 . (
y,
(I_I3 . (x,z)))
by II3Def, D2
;
hence
I_I3 . (
x,
(I_I3 . (y,z)))
= I_I3 . (
y,
(I_I3 . (x,z)))
;
verum end; suppose F1:
(
z = 0 &
y = 0 &
x <> 0 )
;
I_I3 . (x,(I_I3 . (y,z))) = I_I3 . (y,(I_I3 . (x,z)))then
I_I3 . (
y,
z)
= 1
by II3Def;
then I_I3 . (
x,
(I_I3 . (y,z))) =
1
by II3Def
.=
I_I3 . (
y,
(I_I3 . (x,z)))
by F1, II3Def
;
hence
I_I3 . (
x,
(I_I3 . (y,z)))
= I_I3 . (
y,
(I_I3 . (x,z)))
;
verum end; end;
end;
hence
( I_I3 is satisfying_(EP) & not I_I3 is satisfying_(NP) )
by T1, FUZIMPL2:def 2; verum