By Ken Budnick
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We must, however, revise slightly: is the result of replacing (free) the definition of occurrences of s (if s is a variable) by I provided that the variables in s (if s is a term other than a variable) which are bound in A are those variables of I which are bound in A:. Otherwise A: is A. A model for HPC= will be an ordered pair (D, 1) such that: A: A: D is non-empty and I is a function such that I(cn)E D I(pn) c;;, D n I(r) is a function from D n to D. As before we will use sequences a in defining satisfaction and we 38 CHAPTER IV require that a(v) E D, l(r)(a(t l ) ...
Prove this theorem. 36 CHAPTER III THEOREM. There is a decision procedure for formulas of the form (Vt) ... (v n)( 3 vn+t) ... (3 vn+,JB where B contains no quantifiers. EXERCISE 17. Prove this theorem. We have been using the fact that our proof of completeness shows that the system has the subformula property (cf. p. 29), but we can show even stronger results. A is in prenex normal form iff A is of the form (QtVt) (Q2V2) .. (Qnvn)B where each Q is V or 3 and B contains no quantifiers. It can be shown that for any formula C there is a formula A in prenex normal form such that A == C.
The difficulty, of course, is that we need to ensure that all instances of (v)A are T when we conduct our search procedure. The solution then is to keep (v)A in the sequent as a reminder that we haven't finished with it. :1 r where t is not free in EXERCISE 13. backwards. :1 and t is a term. Show that ~ 3 and 3 ~ are sound forwards and Our proof of the completeness of this new system will be like that for GSC, with one complication. In a strict syntactic sense the only subformulas of (x\)F:x\ are itself and F:x\.