Chapter 11 Examples

In order to satisfy the above formula, it is necessary to find an interpretation that satisfies while not satisfying . So take the following:

Clearly is satisfied under this interpretation, since for every natural number , is satisfied. But, is not satisfied, since there are natural numbers that are not even (e.g. 5).

In order to make the above formula true, it is necessary to find an interpretation that does not satisfy or that does satisfy . So take the following:

Now, for this interpretation is not satisfied, since there are natural numbers which are greater than 10 but not even (e.g. 11).

The first formula seems to be a contradictory, since it can never be the case that . So, to show this assume that there is an interpretation (call it ) that satisfies . We know that if satisfies Then for some in the domain satisfies But, then satisfies both and which is a contradiction. So does not satisfy . But since was an arbitrary interpretation we may conclude that every interpretations does not satisfy That is to say, is contradictory.

is a contingent formula. To show this it is necessary to find one interpretation that satisfies it and another interpretation which does not satisfy it.

Take the following interpretation and call it :

In order for the sentence to be satisfied by , it must be the case that satisfies for every and . So, for every and . must not satisfy or satisfy . But this is clearly false, since is satified and is not satisfied.

Take the following interpretation and call it :

In order for the sentence to be satisfied by , it must be the case that satisfies for every and in the domain . So, for every and . must not satisfy or satisfy . But this is clearly true, since for any natural numbers and it is either the case that or

is a tautology. To see this assume that is an arbitrary interpretation and suppose that satisfies . Thus, satisfies for every in the domain. But, then clearly satisfies for every in the domain. Thus, satisfies . So, for any interpretation whenever is satisfied is also satisfied. This is the same as saying that either is not satisfied or is satisfied.

The first argument is valid and can be shown to by valid by the interpretations method as follows:

Assume that an interpretation (call in ) satisfies , and . So first note that satisfies just in case satisfies for some in the domain. Call this term . But now, since we know that satisfies for every in the domain. In particular it holds for , so it must be the case that satisfies and thus satisfies . But we also know that satisfies , in other words does not satisfy or does satisfy . Thus, since we have shown that satisfies above it must be the case that satisfies . So, satisfies for every in the domain, in particular it holds for b. Thus satisfies .

The second argument is invalid and this can be shown by giving a counterexample. A counterexample is simply an interpretation that makes all the premises true while making the conclusion false. So take the following interpretation:

Clearly under this interpretation both and are satisfied. This is true since all natural numbers are either odd or even and there are numbers which are not odd (e.g. 2). But, is clearly false since it is not the case that all natural numbers are even.