# Quantifier proof counterexamples

Providing a counterexample to an invalid quantifier proof is straightforward enough. We need to, unsurprisingly, describe a world in which the premises are true and the conclusion is false. If we do this we have provided a counterexample. The method used in the model answers you may have seen from the past papers may look at first glance technical but it is simply a quick way of describing a world. In describing a world (in which the premises are to be true and the conclusion false) there are essentially two steps – to say what there is in the world, what exists, and then to say what properties and relations obtain in the world. To specify what exists, our domain, we simply list all of the things that exist. Next we specify the extension of all of the properties and relations in our world. We specify the extension of a property by listing all of the things that have it. So we specify the extension of ‘red’ by listing all of the red things. Similarly for relations. So for a two place relation we would list all the pairs of things the relation obtains between. Once this is done for all of the properties necessary then we have described a world. And providing the premises are true in this world and the conclusion false, we have provided a counterexample. Take a look at a question from a couple of years ago.

Premise: ∃x (S(x) ∨ T(x))

Conclusion: ∃x S(x)

This argument is invalid. We ought to be able to determine this from simply understanding what the premise and conclusion mean. The premise is true if there exists something that is S or if there exists something that is T. The conclusion is true just if there exists something that is S. Clearly then it is possible for the premise to be true and the conclusion false. This is because there could be something that was T but nothing that was S. In that case the premise would be true and the conclusion false. This is how we can write this simply:

Domain: { Ayesha }

S(x) : {}

T(x) : { Ayesha }

Firstly the domain is specified. It contains only one thing – Ayesha. So in this world there is but one thing: lonely Ayesha. Then specified is the extension of two predicates, S and T. That there is nothing in the brackets by S means nothing is S. That Ayesha appears in the brackets by T means that Ayesha is T. These three lines provide our counterexample and are all that is required to answer the question.

The following is another example, this time from last years exam.

Again, providing the counterexample is very straightforward – given, that is, that you understand the sentences involved. As you may (/ought to) remember from the course/book, existentially quantified conditionals are easy to mis-read – they are quite an unnatural form of sentence. If we read the first premise as ‘if there exists something that is F then that thing is G’ or ‘if there exists something that is F then there exists something that is G’ we might be tempted to judge the argument valid. Since there exists an F (premise (2)) then must not there exist a G (conclusion) because of premise 1? This would be a mistake.

The most straightforward way to read existentially quantified conditionals is to replace, in your head, the conditional for a disjunction. You will remember from de Morgan’s laws that to do this we simply negate the antecedent (left bit) and swap the arrow for a disjunction symbol. So premise 1 becomes ∃x (¬F(x) OR G(x)). This sentence we can see is true if there is something that is not F and/or there is something that is G. We can now see that the argument is invalid. For the first premise to be true there needs to exists something that is not F and/or something that is G, and for the second premise to be true there needs to exist something that is F. Hopefully we can see that both of these can be true without there existing anything that is G – so the premises can be true and the conclusion false. In the above counterexample we can see that there are just two things in the domain, that a is F and nothing is G. The premises are true in this world because, for premise (1) there is something that is not F (that is, b), and for premise (2) there is something that is F, (that is, a). The conclusion is false in this world for there is nothing that is G.

Question 3 vi from the 06/07 exam is an invalid argument. Have a go at understanding why and providing a counterexample in the above form. To specify the extension of a relation we simply list the pairs (if it’s a two place relation) of things that it holds between. So if there were just four things and one stood in R to another and the other two stood similarly, we would have: domain: {a,b,c,d}, R: {(<a, b>, <c, d>}.

Any last minute questions email me and I will try and help.