I'm fairly certain that this statement is false, but I'm having questions as to how to write it out logically. This is what I have.
(Negation) $\forall$ functions $f$, $g$, $h$, $\in$ $F$, $f \circ h$ = $g \circ h$, but $f \neq$ $g$
Let $x$ be an integer, and suppose:
$f(x) = x$, $g(x) = x^2$, $h(x) = x^0$.
Let x = 3.
Then, $f(3) = 3$, $g(3) = 9$, $h(3) = 1$
$f \circ h(3) = 1$
$g \circ h(3) = 1$.
So, we see that $f \circ h = g \circ h$, but $f \neq g$, So we have proved the original statement to be false.
Is this a valid proof? The part that is concerning me is that im only using one value of $x$, and not showing that for any general value of x, the statement is false. Can anyone clarify is this is logically sound? Thanks in advance.
To say that a set $A$ is equal to set $B$ is to say that both sets contain the same elements. A function is a relation is a set of points. To say that functions $f_1$ and $f_2$ are equal, then, is to say that $f_1$ and $f_2$ contain the same points.
Notice that sets $A=\{1, 2, 3\}$ and $B=\{3, 4, 5\}$ are not equal sets, even though they coincidentally have a common element, $3$.
In order to show that $f\circ h=g\circ h$, we would need to show that all points in set $f\circ h$ are in $g\circ h$ and vice versa. You only showed that one point, namely $(3, 1)$, is in both $f\circ h$ and $g\circ h$. It is just a coincidence.
Had you found an example in which you demonstrate $f\circ h = g\circ h$, that is for all $x$, $(f\circ h)(x) = (g\circ h)(x)$, but $f\ne g$ then you would indeed have disproved the theorem, but, of course, this is not what you did.