# Logic: Problem 3a

### Straightforward solution

Here's one solution to the problem, which conveys the two conditions in a straightforward way:

$$(\exists x \in A,\ P(x)) \ \text{ and } \ (\neg \exists y,z \in A, \ P(y)\ \wedge P(z)\ \wedge\ y \not = z )$$

The parentheses aren't strictly necessary. But they are good to include because they make it easier for a reader to understand the expression.

### Minor variation

Notice that the variable from the first half isn't used in the second half. So it's also correct (though perhaps a bit less clear) to re-use the variable name x, as follows:

$$(\exists x \in A,\ P(x)) \ \text{ and } \ (\neg \exists x,y \in A, \ P(x)\ \wedge P(y)\ \wedge\ x \not = y )$$

Notice that the variable x in the first half and the variable x in the second half are two distinct variables.

Here's a rather different sort of answer, equivalent to the above. It's a more hard-core mathematician style of answer: concise but harder to understand. $$\exists x \in A, \ P(x)\ \text{ and } \ ( \forall y \in A,\ P(y) \rightarrow x = y )$$
$$\exists x \in A,\ P(x)\ \text{ and } \ (\neg \exists y \in A, \ P(y)\ \wedge\ x \not = y )$$