Here we will prove the following statement.

Let f, g be real functions. If they are odd, then f⋅g is an even function.

We know that when translated into formal logical statement, this reads:

For every real function f and for every real function g:
If ( f is odd and g is odd), then f ⋅g is even.

As usual, the general quantifier means that we take arbitrary two real functions f, g about which we do not know anything else yet and show that the implication is satified for them. We will try a direct proof, so we now also assume that these two functions are odd and see whether their product h = f ⋅g is an even function. How do we show this? Definition is usually a good start.

How do we recognize whether the function h is even? We take some x from its domain (arbitrary! since in the definition of even function there is a general quantifier), then substitute -x into h and see what happens. If we can transform this expression back to h(x), we have ourselves an even function.

Now we apply this to h = f ⋅g. Actually, is the number -x in the domain? Well, yes. The domain of f ⋅g is given as the intersection of domains of f and g, so x is in D( f ) and D(g). But since these two functions are symmetric, then also -x must lie in these two domains, hence it lies in their intersection, which is the domain of h. Thus we can substitute -x into h. A mathematician has to be careful about all the details and check that there is no hidden flaw in the argument.

So we know that we can substitute. What do we know about h(−x)? By the definition of product of functions it is f (−x)⋅g(−x), do we have any information about f and g applied to -x? Actually we do, now we can use our assumption that f and g are odd. This looks very promising, we try to finish this calculation:

h(−x) = f (−x)⋅g(−x) = [−f (x)]⋅[−g(x)] = f (x)⋅g(x) = h(x).

We just proved that for every x from D(h) we have h(−x) = h(x), which by definition shows that h is an even function. Since f and g were arbitrary odd functions in this argument, the given statement is proved.

Note that we used the assumption in our proof. This is to be expected. If you try to prove some implication and manage to do so without using its assumption, then most likely you missed something.

Note also that the definition of symmetry played two different roles in our proof. Whether a function is even or odd can be determined using a test, a condition we will now call C. When we wanted to show that h is even, we needed to show that C is satisfied. At that moment validity of C was our goal. But when we worked with f and g, we knew that they were odd, so then we also knew that the corresponding condition C must be satisfied and therefore we were allowed to use it as a fact. Thus in definition, the condition C serves as a test when we want to decide on a property, or as something that is available to us in case we already know that the property si true.