Constructing a model universe
We now show how to determine validity and other
semantic properties in QL by constructing a model universe. A model universe is
simply a domain of the appropriate size for determining a semantic property. We
will discuss how to determine the appropriate size of the domain below.
Let's begin with a syllogism, because we already have a test for the validity of syllogisms.
| All philosophers are bald. |
| Anyone who is bald is foolish. |
| Some philosophers are foolish. |
You should already be able to tell that this is invalid by inspection. But if you can't, go ahead and do a Venn Diagram.
How do we test this for validity using the semantics of QL? First we translated it into QL:
| (x)(Px ⊃ Bx) |
| (x)(Bx ⊃ Fx) |
| (∃x)(Px & Fx) |
Now we choose a domain. But how many individuals should our domain have? Well, the smaller the domain, the easier it will be to test. So let's start with a domain of one individual, {a}. (Note: We could have started with the domain {}. What would we conclude in that domain?) We then rewrite our argument as:
| (Pa ⊃ Ba) |
| (Ba ⊃ Fa) |
| (Pa & Fa) |
and we do a truth-table:
| Pa | Ba | Fa | (Pa ⊃ Ba) | (Ba ⊃ Fa) | (Pa & Fa) | |
| T | T | T | T | T | T | |
| T | T | F | T | F | F | |
| T | F | T | F | T | T | |
| T | F | F | F | T | F | |
| F | T | T | T | T | F | |
| F | T | F | T | F | F | |
| F | F | T | T | T | F | |
| F | F | F | T | T | F |
The truth-table shows that in a domain of one individual, the argument is invalid. Remember that an argument is invalid in QL just in case there is any domain where it is has true premises and a false conclusion under one interpretion, that is on one row of its truth-table.
Suppose that the argument didn't come out invalid in a domain of one individual. Could we conclude that it is valid? The answer, importantly, is "no". The argument is valid just in case there is no domain where the truth-table has an invaliditing row. So do you have to keep checking every domain? Fortunately, if an argument is invalid in a domain, it will be invalid in a domain no greater than in a domain of 2n individuals, where n is the number of distinct propositional functions. So the largest domain we would need to look for with the argument above is a domain of 8 individuals. (In the worst case scenario, then we would need a truth table with 28 or 256 rows! Fortunately, we'll see that we will soon introduce truth-tree method, which is much more efficient. So we'll never have to compute such lengthy truth-tables).
It will be helpful to look at another argument, which is also invalid, but not in a domain of one individual:
| All mice are four-legged |
| Something isn't four-legged |
| There are no mice. |
This is clearly also an invalid argument. It could certainly be true that all mice are four-legged, and that something (a tomato, say) is not four legged, and false that there are no mice. But to show this, a domain of one won't do:
| (x)(Mx ⊃ Fx) |
| (∃x)~Fx |
| ~(∃x)Mx |
In order to rewrite this argument in a model universe, we need to convert the conclusion to a quantified proposition, using the Quantifier Negation rules we encountered earlier. Changing the conclusion fro the negation of an existentially quantified wff to a universally quantified negation we have:
| (x)(Mx ⊃ Fx) |
| (∃x)~Fx |
| (x)~Mx |
Now let's rewrite the argument in a model universe of one individual {a}:
| (Ma ⊃ Fa) |
| ~Fa |
| ~Ma |
Do a truth-table for this argument and you will see that it is valid. (Note that it has the form of the rule of inference, modus tollens.) So let's try a domain of two individuals, {a,b}. Remember that we have to turn universally quantified wffs into conjunctions and existentially quantified wffs into disjunctions:
| (Ma ⊃ Fa) & (Mb ⊃ Fb) |
| ~Fa v ~Fb |
| ~Ma & ~Mb |
There are four distinct singular propositions here, so we will need a truth-table with 16 rows! Let's do it - though we'll go gappy and only look at rows where the conclusion comes out false, stopping when we come to an invalidating row.
| Ma | Fa | Mb | Fb | (Ma ⊃ Fa) & (Mb ⊃ Fb) | ~Fa v ~Fb | ~Ma & ~Mb | |
| T | T | T | T | F | F | ||
| T | T | T | F | F | T | F | |
| T | T | F | T | F | |||
| T | T | F | F | T | T | F | |
| T | F | T | T | F | |||
| T | F | T | F | F | |||
| T | F | F | T | F | |||
| T | F | F | F | F | |||
| F | T | T | T | F | |||
| F | T | T | F | F | |||
| F | T | F | T | ||||
| F | T | F | F | ||||
| F | F | T | T | F | |||
| F | F | T | F | F | |||
| F | F | F | T | ||||
| F | F | F | F |
Once we computed the fourth row, we found that the argument is invalid. It should be obvious how this method applies to all the semantic properties of QL. Here is an exercise to do using this method. We now turn to using truth-tress in QL, which makes life much easier!
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