I would like to define a pseudo-boolean function $f$ such that $f(x) = 0$ for all logically valid combinations of $x\in{0,1}$ and $f(x) > 0$ for all logically invalid combinations of $x\in{0,1}$.
Suppose I have the following constraint where $x_i \in {0,1}$:
$x_1 + x_2 + x_3 \leq 2$
| $x_1$ | $x_2$ | $x_3$ | Valid |
|---|---|---|---|
| 0 | 0 | 0 | Yes |
| 0 | 0 | 1 | Yes |
| 0 | 1 | 0 | Yes |
| 0 | 1 | 1 | Yes |
| 1 | 0 | 0 | Yes |
| 1 | 0 | 1 | Yes |
| 1 | 1 | 0 | Yes |
| 1 | 1 | 1 | No |
After binary expansion, it becomes quite trivial to see that the case $\{1,1,1\}$ is the only case that causes an invalid result. As such, $f = x_1 \land x_2 \land x_3$ is an appropriate pseudo boolean function.
Is there an efficient way to reach this conclusion without a binary expansion?
Alternatively, I can surmise that the "invalid" cases are:
$x_1 + x_2 + x_3 > 1$
Since we are dealing with binary values, this is equivalent to:
$x_1 + x_2 + x_3 \geq 2$
That looks a little bit closer to $x_1 \land x_2 \land x_3$, but I'm still not sure...
