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Theorem 20
φ is a theorem of WMTL . Thus
WMTL , differently from IPC , satisfies the De Morgan laws .
The weak excluded middle
¬
φ
∨¬¬
Ob DB sk , ¬
⊕¬¬
Proof We have to show that for each A
=
TA
A
A
=
Υ , that is,
T( ¬
∪¬¬
for each (also infinite) relation R
A) .
Hence J ¬ A = J Υ \ J A = J Υ \ J ¬¬ A , so that at least one of J ¬ A and J ¬¬ A = J A
has to be finite. Let us consider the case when J ¬¬ A = J A is finite. In what follows,
the values in J A will be denoted by a i while the values in J ¬ A will be denoted by x i .
Thus, R
Υ , R
A
R 2 , where R 1 in each tuple has at least one value a i , while R 2
has only the tuples with values x i in the infinite set J ¬ A , so that
1. R 2 ∈¬
=
R 1
A can be an infinite relation.
R 1 can be an infinite relation as well because it has not only the values a i of the
finite set J A (in every its tuple) but also the values x i of the (possibly infinite) set
J ¬ A .
Let us take any tuple t I
1 of finite
substrings (the arity of each relation is always finite) composed of only values in
J A : we denote by i m
R 1 composed of the finite number n
=
1 ,...,n , the position of the m th substring of a length
k m , composed of only elements in J A (so that i m i l iff m l ), and set i 0 =
1, m
1,
k 0 =
1.
The remaining part of this tuple t I is composed of substrings with the values x i
in J ¬ A . We can have at maximum n
l 0 =
0, and i n + 1 =
ar(R)
+
+
1 such substrings, each one of the length
l m + 1 =
0 , 1 ,...,n .
The I is a complete ordered set of indexes of values a i in this tuple t I , thus equal
i m + 1
(i m +
k m ) , m
=
to:
I
=
i 1 ,...,i 1 +
k 1
1 ,i 2 ,...,i 2 +
k 2
1 ,...,i n ,...,i n +
k n
1
.
The complete ordered set of indexes of values x i in J ¬ A in this tuple t I is then
J
=
i m +
k m ,...,i m + 1
1 ,i m + 1 +
k m + 1 ,...,i m + 2
1 ,,...,i l +
k l ,...,i l + 1
1
,
where m
=
0if i 1 > 1; 1 otherwise, and l
=
n if i n +
k n <i n + 1 =
ar(R)
+
1; n
1
otherwise.
Let us consider, for example, the tuple
t I =
x 1 x 2 x 3 a 1 a 2 x 4 x 5 a 3 a 4 a 5 x 6 a 6 x 7 x 8 x 9
,
so that n
=
3; i 1 =
4 ,i 2 =
8 ,i 3 =
12 ,i 4 =
ar(R)
+
1
=
16; k 1 =
2 ,k 2 =
3 ,k 3 =
1;
and l 1 =
3 ,l 2 =
2 ,l 3 =
1 ,l 4 =
3, and hence,
I =
4 , 5 , 8 , 9 , 10 , 12
and J =
1 , 2 , 3 , 6 , 7 , 11 , 13 , 14 , 15
.
Let us define a function f
:{
1 ,...,n
}→ N
such that f( 1 )
=
0 and f(l)
=
1 j l 1 k j ,for2
l
n . Then we define the relation:
a f(l) + m + 1 ) ,
R 1 WHERE
1
R I =
(nome i l + m =
l
n
0
m
k l
1
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