Math 470 Spring ’05
HW #8 Solutions
Section 2.4
9. If  is quantifier-free and contains no free variables, it is a ground sentence. In any interpretation its truth or falsity is determined by the truth or falsity of the atomic formulas contained in it, so when each atomic formula R(t1, …, tn) is replaced by a propositional letter to obtain  ’, the truth value of  can be determined from the truth value of the atomic formulas in it by truth table. Then  is valid iff  ’ is valid.
10. Since  (x1…xn) and  (c1…cn) have the same propositional structure, the same  ’ is obtained from either of them.
By exercise 8,  (x1, …, xn) is valid iff  (c1, …. ,cn) is valid and by exercise 9,
 (c1…cn) is valid iff  ’ is valid.
Thus by transitivity,  (x1… xn) is valid iff  ’ is.
Section 2.6
13.
Suppose that we have a CST giving a tableau proof T of  (c1…cn). This is a contradictory tableau with root F  (c1…cn). Form a new tableau with root
F  x1…  xn  (x1…xn) . Apply Rule 7b n times to get F  (c1…cn) (since the ci are not in L). The result is a contradictory tableau that is a tableau proof of  x1…  xn
 (x1…xn).
On the other hand, suppose that we have a contradictory CST with root
F  x1…  xn  (x1…xn) . This tableau must begin with n applications of Rule 7b. The new constants can be replaced by c1… cn, since these are not in L. The remainder of the tableau is a proof of  (c1…cn).
Section 2.10
1,. a) {c}
b) R(c) is true or R(c) is false. c) S is satisfiable. Take A = N, c = 0 and R(x) = (x = 0),i.e., R is true for ) and false for all other numbers. Than A satisfies S.
S has no Herbrand model because the Herbrand universe has only one element and R is either true or false for that element.
The situation can be rectified by Skolemizing  x  R(x) to  R(d).
Then the Herbrand universe is {c, d} and S has a Herbrand model: R(c) is true and
R(d) is false.
2. a) {c, fc, ffc, fffc, …) b) Let Rn be true for terms of length <=n and false for terms of length >n.
There are infinitely many such R and each gives a different Herbrand structure.
6. By Exercise 6.13,  is valid iff  y1…  ym  (c1…cn, y1…ym) is valid. By
Corollary 10.5, this is valid iff there are ground terms t1,1…t1,m … tn,1 … tn,m such that  (t1,1…t1,m) v …  (tn,1…tn,m) is a tautology.

Since the language has no function symbols, the Herbrand universe consists only of constants. Thus the ground terms are only these constants, say di. If there are finitely many of them, the formula is valid iff the disjunction of all combinations of the di is valid.
If there are infinitely many, they can be checked sequentially. This procedure will continue indefinitely if the formula is not valid.