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# Let $f,\, g\, and\, h$ be functions from $R\, to\, R.$ Show that $(f . g) oh = (foh) . (goh)$

Toolbox:
• Given two functions $f:A \to B$ and $g:B \to C$, then composition of $f$ and $g$, $gof:A \to C$ by $gof (x)=g(f(x))\;for\; all \;x \in A$
We have to prove: L.H.S.: $(f.g) oh =$ R.H.S.: $(foh ).(goh)$.
Consider L.H.S. $(f.g) oh$,
$\Rightarrow (f.g) oh = (f.g) o (h(x))$
$\Rightarrow (f.g) oh = f(h(x)) \times g(h(x))$
We know that given two functions $f:A \to B$ and $g:B \to C$, then composition of $f$ and $g$, $gof:A \to C$ by $gof (x)=g(f(x))\;for\; all \;x \in A$
$\Rightarrow f(h(x)) = foh$ and $g(h(x)) = goh$
$\Rightarrow$ L.H.S. $= f(h(x)) \times g(h(x)) = (foh) .(goh)$ = R.H.S.