84
Mathematical Aspects of Logic Programming Semantics
(b) Take ≤ to be either ≤ k or ≤ t on the logic FOU R. Form the domain
I(X, T ) with the corresponding pointwise order [ and the corresponding
product domain I(X, T ) × I(X, T ). Then both ∨ and ∧ are Scott continuous as mappings from I(X, T )×I(X, T ) to I(X, T ). The same statements
are true in the case of classical two-valued logic (where the ordering has
to be ≤ t ) and Kleene’s strong three-valued logic.
Proof: For (a), the statements concerning the truth ordering [ t have already
been established. To deal with [ k , we use the criteria for convergence presented in Theorem 3.2.6. Let (v i ) be a net converging in the Scott topology to
v in I(X, T ). Suppose that x ∈ (¬v) t . Then (¬v)(x) = t, and hence x ∈ v f .
Since v i → v, we have that eventually x ∈ v i f ∪v i b , that is, eventually v i (x) = f
or v i (x) = b. But then eventually ¬v i (x) = t or ¬v i (x) = b, and so eventually
x ∈ (¬v i ) t ∪ (¬v i ) b . The other cases are handled similarly. Thus, the net (¬v i )
converges to ¬v in the Scott topology, as required.
For (b), we establish the result stated concerning ∨, noting that the proof
for ∧ is entirely similar. Now, as is well-known, it suffices to show continuity
in each argument
15 of ∨, and, by commutativity, it in fact suffices to show
continuity in one argument, the first, say. So, fix v ∈ I(X, T ), and suppose
that u i → u in the Scott topology on I(X, T ). Let x ∈ X be arbitrary. Then
(u ∨ v)(x) = u(x) ∨ v(x). Since u i → u, we have eventually that u(x) ≤ u i (x)
by Theorem 3.2.6. Therefore, by Proposition 1.3.7, we have eventually that
u(x) ∨ v(x) ≤ u i (x) ∨ v(x), and this suffices, by Theorem 3.2.6, to show that
u i ∨ v → u ∨ v, as required.
•
We now turn our attention to these same operators in relation to the
topology Q. Indeed, we close this chapter with the following result.
3.4.2 Theorem Let T denote Belnap’s logic FOU R. Then the following
statements hold.
(a) The negation operator ¬ : I(X, T ) → I(X, T ) is continuous in the topology Q. Hence, it is continuous in Q when T denotes either classical twovalued logic or Kleene’s strong three-valued logic.
(b) Both ∨ and ∧ are continuous as mappings from I(X, T ) × I(X, T ) to
I(X, T ), where I(X, T ) × I(X, T ) is endowed with the product topology
of Q with itself. Hence, the same result holds relative to either classical
two-valued logic or Kleene’s strong three-valued logic.
Proof: For (a), let (v i ) be a net converging to v in I(X, T ) relative to the
topology Q, and let x ∈ X be arbitrary. Then eventually v i (x) = v(x). Therefore, eventually (¬v i )(x) = (¬v)(x). Therefore, ¬v i → ¬v in Q, and the result
follows.
For (b), let (u i , v i ) → (u, v) in the product topology. Then u i → u in Q
15 See Proposition 2.4 of [Stoltenberg-Hansen et al., 1994].
Mathematical Aspects of Logic Programming Semantics
(b) Take ≤ to be either ≤ k or ≤ t on the logic FOU R. Form the domain
I(X, T ) with the corresponding pointwise order [ and the corresponding
product domain I(X, T ) × I(X, T ). Then both ∨ and ∧ are Scott continuous as mappings from I(X, T )×I(X, T ) to I(X, T ). The same statements
are true in the case of classical two-valued logic (where the ordering has
to be ≤ t ) and Kleene’s strong three-valued logic.
Proof: For (a), the statements concerning the truth ordering [ t have already
been established. To deal with [ k , we use the criteria for convergence presented in Theorem 3.2.6. Let (v i ) be a net converging in the Scott topology to
v in I(X, T ). Suppose that x ∈ (¬v) t . Then (¬v)(x) = t, and hence x ∈ v f .
Since v i → v, we have that eventually x ∈ v i f ∪v i b , that is, eventually v i (x) = f
or v i (x) = b. But then eventually ¬v i (x) = t or ¬v i (x) = b, and so eventually
x ∈ (¬v i ) t ∪ (¬v i ) b . The other cases are handled similarly. Thus, the net (¬v i )
converges to ¬v in the Scott topology, as required.
For (b), we establish the result stated concerning ∨, noting that the proof
for ∧ is entirely similar. Now, as is well-known, it suffices to show continuity
in each argument
15 of ∨, and, by commutativity, it in fact suffices to show
continuity in one argument, the first, say. So, fix v ∈ I(X, T ), and suppose
that u i → u in the Scott topology on I(X, T ). Let x ∈ X be arbitrary. Then
(u ∨ v)(x) = u(x) ∨ v(x). Since u i → u, we have eventually that u(x) ≤ u i (x)
by Theorem 3.2.6. Therefore, by Proposition 1.3.7, we have eventually that
u(x) ∨ v(x) ≤ u i (x) ∨ v(x), and this suffices, by Theorem 3.2.6, to show that
u i ∨ v → u ∨ v, as required.
•
We now turn our attention to these same operators in relation to the
topology Q. Indeed, we close this chapter with the following result.
3.4.2 Theorem Let T denote Belnap’s logic FOU R. Then the following
statements hold.
(a) The negation operator ¬ : I(X, T ) → I(X, T ) is continuous in the topology Q. Hence, it is continuous in Q when T denotes either classical twovalued logic or Kleene’s strong three-valued logic.
(b) Both ∨ and ∧ are continuous as mappings from I(X, T ) × I(X, T ) to
I(X, T ), where I(X, T ) × I(X, T ) is endowed with the product topology
of Q with itself. Hence, the same result holds relative to either classical
two-valued logic or Kleene’s strong three-valued logic.
Proof: For (a), let (v i ) be a net converging to v in I(X, T ) relative to the
topology Q, and let x ∈ X be arbitrary. Then eventually v i (x) = v(x). Therefore, eventually (¬v i )(x) = (¬v)(x). Therefore, ¬v i → ¬v in Q, and the result
follows.
For (b), let (u i , v i ) → (u, v) in the product topology. Then u i → u in Q
15 See Proposition 2.4 of [Stoltenberg-Hansen et al., 1994].