Proposition 10 For every program P and procedure call X in P, P^ is
continuous in X.
Proof. By induction on the complexity of P, using the previous lemma for
the case that P = hXi ^ Piii(Q). □
We can prove a similar connection between correctness formulae and weak
anticontinuity. First we note that there are certain syntactic features which
will guarantee a type logical term to be continuous or anticontinuous in a
given variable.
Lemma 11
1. If φ is continuous in ξ then -φ is anticontinuous in ξ;
2. If φ is anticontinuous in ξ then -φ is continuous in ξ;
3. If φ,ψ are (anti-)continuous in ξ then φ ∧ ψ is (anti-)continuous in ξ;
4. If Aα→t is (anti-)continuous in ξ and ξ does not occur in δα then A(δ)
is (anti-)continuous in ξ;
5. If φ is (anti-)continuous in ξ then λζα.φ is (anti-)continuous in ξ;
6. If φ is anticontinuous in ξ then ∀ζ φ is anticontinuous in ξ;
7. If φ is continuous in ξ then ∃ζ φ is continuous in ξ.
Proof. By inspection of the various cases. To prove the first case, for example,
assume that φ is continuous in ξ, and let R0 ⊆ R1 ⊆ ... ⊆ Rn ⊆ ... be a
chain. Then Tn V(-φ,a[Rn/ξ]) = 1 iff, for all n, V(-φ, a[Rn∕ξ]) = 1 iff (by
Definition 10), for all n, V(φ, a[Rn∕ξ]) = 0 iff Sn V(φ, a[Rn∕ξ]) = 0 iff (by
continuity of φ) V(φ, a[Rω/ξ]) = 0 iff (by Definition 10) V(-φ, a[Rω/ξ]) = 1.
The other cases are equally simple and are left to the reader. □
The relation between correctness formulas and weak anticontinuity that was
promised is stated and proved as follows.
Proposition 12 For every correctness formula C and procedure call X in
C, C t is weakly anticontinuous in X .
22
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