# Is there one anti-model that refutes three invalid inferences based on “more than half"?

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?

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.