Let $f, g$ and $h$ be functions from $R$ to $R$. Show that:
$\begin{cases} (f+g)oh = foh + goh \\ (f \cdot g)oh = (foh) \cdot (goh) \end{cases}$
$\begin{cases} (f+g)oh = foh + goh \\ (f \cdot g)oh = (foh) \cdot (goh) \end{cases}$
To prove: $(f+g)oh = foh + goh$
$LHS = [(f+g)oh](x)$
$= (f+g)[h(x)] = f[h(x)] + g[h(x)]$
$= (foh)(x) + (goh)(x)$
$= \{(foh) + (goh)\}(x) = RHS$
$\therefore \{(f+g)oh\}(x) = \{(foh) + (goh)\}(x)$ for all $x \in R$.
Hence,$(f+g)oh = foh + goh$.
To prove: $(f \cdot g)oh = (foh) \cdot (goh)$
$LHS = [(f \cdot g)oh](x)$
$= (f \cdot g)[h(x)] = f[h(x)] \cdot g[h(x)]$
$= (foh)(x) \cdot (goh)(x)$
$= \{(foh) \cdot (goh)\}(x) = RHS$
$\therefore [(f \cdot g)oh](x) = \{(foh) \cdot (goh)\}(x)$ for all $x \in R$.
Hence,$(f \cdot g)oh = (foh) \cdot (goh)$.
$LHS = [(f+g)oh](x)$
$= (f+g)[h(x)] = f[h(x)] + g[h(x)]$
$= (foh)(x) + (goh)(x)$
$= \{(foh) + (goh)\}(x) = RHS$
$\therefore \{(f+g)oh\}(x) = \{(foh) + (goh)\}(x)$ for all $x \in R$.
Hence,$(f+g)oh = foh + goh$.
To prove: $(f \cdot g)oh = (foh) \cdot (goh)$
$LHS = [(f \cdot g)oh](x)$
$= (f \cdot g)[h(x)] = f[h(x)] \cdot g[h(x)]$
$= (foh)(x) \cdot (goh)(x)$
$= \{(foh) \cdot (goh)\}(x) = RHS$
$\therefore [(f \cdot g)oh](x) = \{(foh) \cdot (goh)\}(x)$ for all $x \in R$.
Hence,$(f \cdot g)oh = (foh) \cdot (goh)$.
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