From: Philip Johnson-Laird BA PhD Psychology (UCL), Stuart Professor of Psychology Emeritus at Princeton. (Author isn't a logician.) How We Reason (1st edn 2008). p. 224-225.

Is there a single anti-model that can refute 1-3 beneath?

enter image description here


It is not possible.

Let $n$ be the total number of people. Then we have: $$N_{abc}+N_{ab}+N_{ac}+N_{bc}+N_a+N_b+N_c \le n$$ where each $N_*$ represents the number of people with exactly the jobs in the subscript. It is an inequality since there may be people who are none of the three.

Lets define $N_2 = N_{ab}+N_{bc}+N_{ab}$, and $N_1 = N_a+N_b+N_c$. The above inequality is then:
$$N_{abc}+N_2+N_1 \le n$$ We are given that more than half the people have each job: $$N_{abc}+N_{ab}+N_{ac}+N_a > \frac{n}{2}\\N_{abc}+N_{ab}+N_{bc}+N_b > \frac{n}{2}\\N_{abc}+N_{ac}+N_{bc}+N_c > \frac{n}{2}$$ Adding these three together we get: $$ 3N_{abc}+2N_2+N_1 > \frac{3n}{2} $$ To refute conclusion 2 (and 1) we need $N_{abc}=0$. The two inqualities then become: $$N_2+N_1 \le n \\ 2N_2+N_1 > \frac{3n}{2} $$ If we negate the first inequality (which flips its direction) and add it to the second, we get $N_2 > \frac{n}{2}$.

To refute conclusion 3 we need $N_2 \le \frac{n}{2}$, so it is not possible to refute all conclusions with a single counter-example.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.