# Maths and Programs

Let’s solve this question (taken from the second assignment of Functional Programming in Scala):

Using

`forall`

, implement a function`exists`

which tests whether a set contains at least one element for which the given predicate is true. Note that the functions`forall`

and`exists`

behave like the universal and existential quantifiers of first-order logic.

The code listing for `forall`

is as follows:

```
def forall(s: Set, p: Int => Boolean): Boolean = {
def iter(a: Int): Boolean = {
if (a > bound) true
else if (contains(s, a) && !p(a)) false
else iter(a + 1)
}
iter(-bound)
}
```

where `bound`

is `1000`

.

A set is a defined as follows, along with the predicate to test for containment.

Here is how I arrive at the answer for the question:

- From the question, it is intuitive to try to stop the iteration when there is an element that satisfies the predicate
`p`

, which is exactly what`exists`

should do.- The iteration is stopped only when the current element
`a`

does not satisfy`p`

.- However, we need to satisfy
`p`

in order for`exists`

to correctly do its job.

In conclusion, we need a double negation here. In other words, we need to negate

`p`

and then negate the answer returned by`forall`

when called with the negation of`p`

.

From that reasoning, it is easy to code up the first attempt at a solution.

Now, let’s see how this solution fares.

```
scala> def natural = (x: Int) => x > 0
scala> forall(natural, x % 2 == 0) // should return false, as not all naturals are even
Res01 = false
scala> exists(natural, x % 2 == 0) // should return true
Res02 = true
```

The solution passes the tests. Let’s look at how to explain the code simply. The code says

There exists in set

`s`

an element that satisfies the predicate`p`

when not all of the elements in`s`

satisfies the negation of`p`

.

That makes sense. But lo and behold, the code reflects a mathematical truth, and it even looks like it is translated verbatim from mathematical symbols to programmatic symbols!

∃x:P(x) ⇔ ¬∀x:¬P(x)

compared with

The correspondence is remarkable!

After reading this post, you should try this question. The answer will pleasantly surprise you.

Finally, write a function

`map`

which transforms a given set into another one by applying to each of its element the given function.`map`

has the following signature: