Introduction to Symbolic Logic: 'Understanding Symbolic Logic, 2nd Edition,' by Virginia Klenk, Page 294

Question

I read this passage in my textbook:

...if there is a counterexample in a domain with $m$ individuals, then there is also a counterexample in all larger domains. It follows by contraposition (and C.Q.N.) that if there is no counterexample in a domain with m individuals, then there is also no counterexample in any smaller domain.Thus if there is no counterexample in a domain with...[m]...individuals...then there can be no counterexample in any smaller domain or in any larger domain...

I've been trying for about an hour now to symbolize and prove this passage's argument, but this is all I can get:

Cx = x is a counter-example, Mx = x is in a domain with m individuals, Lx = x is in a domain with larger-than-m individuals

Premise: If (∃x)(Cx & Mx), then (∀x)(If Lx, then Cx). Conclusion: If ~(∃x)(Cx & Mx), then ~(∃x)(Cx & ~Lx)

The textbook passage argues that you can get to that conclusion from that premise via contraposition and the categorical quantifier negation rules, but I can't figure out how to do that.

Could someone help? Thank you so much!

Answer 1

You have slightly altered the quotation, omitting the context...

I'm refeering to :

• Virginia Klenk, Understanding Symbolic Logic (Fifth ed, 2008), page 310.

There are two facts in place here.

One is the fact that, for monadic predicate logic :

[if] $n$ is the number of different predicate letters in an argument form, the largest domain you need to test is one with $2^n$ individuals.

If there is no counterexample in this domain, then there is no counterexample in any larger domain; thus, if you have tested all domains up to that point, you may conclude that the argument form is valid.

We can translate it in symbols with a "complex" predicate $Counter(x,n)$ meaning "$x$ is a counterexample in a domain with $n$ individuals" :

$\lnot \ \exists x \ Counter(x,2^n) \to \forall k \ [(k \ge 2^n) \to \lnot \ \exists y \ Counter(y,k)]$ --- (*).

Next, we have a second fact, holding in general (i.e. for a formula of predicate calculus whatever) :

if there is a counterexample in a domain with $m$ individuals, then there is also a counterexample in all larger domains.

In symbols :

$\exists x \ Counter(x,m) \to \forall k \ [(k > m) \to \exists y \ Counter(y,k)]$ --- (§).

Assume now that there is no counterexample in a domain with $m$ individuals; if we have a counterexample in a domain with less than $m$ individuals, say, $k < m$, by the above we would have a counterexample also in all larger domains, ans so also in a domain with $m$ individuals, because $m$ is larger than $k$: contradiction.

The argument assumes :

1) $\lnot \exists x \ Counter(x,m)$

2) $\exists k_0 \ [\exists y \ Counter(y,k_0) \land (k_0 < m)]$.

From 2) it follows that : $\exists y \ Counter(y,k_0)$ and with (§) we have that :

$\forall k \ [(k > k_0) \to \exists y \ Counter(y,k)]$.

But by 2) we have also : $m > k_0$; thus : $\exists y \ Counter(y,m)$, contradicting 1).

Thus, form 1) it follows the negation of 2), i.e. :

$\lnot \exists x \ Counter(x,m) \to \forall k \ \lnot \ [\exists y \ Counter(y,k) \land (k < m)]$.

But $\lnot (p \land q)$ is equivalent to : $q \to \lnot p$; thus :

$\lnot \ \exists x \ Counter(x,m) \to \forall k \ [(k < m) \to \lnot \ \exists y \ Counter(y,k)]$.

This holds in general; so, also for $2^n$ :

$\lnot \ \exists x \ Counter(x,2^n) \to \forall k \ [(k < 2^n) \to \lnot \ \exists y \ Counter(y,k)]$ --- (@).

Now the result follows from (*) and (@) and Proof by cases (or Disjunction elimination) with the fact that :

$(k < 2^n) \lor (k \ge 2^n)$.