Consider $f: N \rightarrow N$,$g: N \rightarrow N$,and $h: N \rightarrow R$ defined as $f(x) = 2x$,$g(y) = 3y + 4$,and $h(z) = \sin z$,$\forall x, y, z \in N$. Show that $h \circ (g \circ f) = (h \circ g) \circ f$.

(N/A) We have:
$h \circ (g \circ f)(x) = h(g(f(x)))$
$= h(g(2x))$
$= h(3(2x) + 4)$
$= h(6x + 4)$
$= \sin(6x + 4), \forall x \in N$
Also,$((h \circ g) \circ f)(x) = (h \circ g)(f(x))$
$= (h \circ g)(2x)$
$= h(g(2x))$
$= h(3(2x) + 4)$
$= h(6x + 4)$
$= \sin(6x + 4), \forall x \in N$
Since both expressions yield the same result,it is shown that $h \circ (g \circ f) = (h \circ g) \circ f$.
This result holds true in general for the composition of functions.