To take an example, here is the way I may argue for the first of the formulas. Given any interpretation, consider the set corresponding to P under it. If this set includes the entire domain, then (all x) P(x) must be satisfied by any interpretation and it quickly follows that the entire formula is satisfied. On the other hand, suppose that the interpretation of P does not include at least one element from the domain. Let us call this a. Then the interpretation together with the assignment that maps x to a satisfies the formula (P(x) => (all x) P(x)). But then it follows that the given formula is satisfied by the interpretation.
As a particular application of the last problem, consider showing the following: A formula F in negation normal form that does not contain existential quantifiers or the equality symbol is unsatisfiable if and only if it is unsatisfiable in the term model. This observation is central to the refutation approach to theorem proving.
Last updated on Nov 1, 2004 by firstname.lastname@example.org.