set I = I_I4 ;
ex y being Element of [.0,1.] st I_I4 . (1,y) <> y
then T1:
not I_I4 is satisfying_(NP)
by FUZIMPL2:def 1;
for x, y, z being Element of [.0,1.] holds I_I4 . (x,(I_I4 . (y,z))) = I_I4 . (y,(I_I4 . (x,z)))
proof
let x,
y,
z be
Element of
[.0,1.];
I_I4 . (x,(I_I4 . (y,z))) = I_I4 . (y,(I_I4 . (x,z)))
per cases
( z = 1 or ( z <> 1 & y <> 1 & x <> 1 ) or ( z <> 1 & y <> 1 & x = 1 ) or ( z <> 1 & y = 1 & x = 1 ) or ( z <> 1 & y = 1 & x <> 1 ) )
;
suppose F1:
(
z <> 1 &
y <> 1 &
x = 1 )
;
I_I4 . (x,(I_I4 . (y,z))) = I_I4 . (y,(I_I4 . (x,z)))then
(
I_I4 . (
y,
z)
= 1 &
I_I4 . (
x,
z)
= 0 )
by II4Def;
then I_I4 . (
x,
(I_I4 . (y,z))) =
1
by II4Def
.=
I_I4 . (
y,
(I_I4 . (x,z)))
by F1, II4Def
;
hence
I_I4 . (
x,
(I_I4 . (y,z)))
= I_I4 . (
y,
(I_I4 . (x,z)))
;
verum end; suppose F1:
(
z <> 1 &
y = 1 &
x <> 1 )
;
I_I4 . (x,(I_I4 . (y,z))) = I_I4 . (y,(I_I4 . (x,z)))then F3:
I_I4 . (
x,
z)
= 1
by II4Def;
I_I4 . (
x,
(I_I4 . (y,z))) =
1
by II4Def, F1
.=
I_I4 . (
y,
(I_I4 . (x,z)))
by F3, II4Def
;
hence
I_I4 . (
x,
(I_I4 . (y,z)))
= I_I4 . (
y,
(I_I4 . (x,z)))
;
verum end; end;
end;
hence
( I_I4 is satisfying_(EP) & not I_I4 is satisfying_(NP) )
by T1, FUZIMPL2:def 2; verum