Showing $\bar{p} * \alpha * p \simeq_{x_0} \bar{p} * \beta * p$, path-homotopy For given path-connected topological spaces $X$, $x_0 , x_1 \in X$, and given loops $\alpha$, $\beta$ $: I \rightarrow X$ with base point at $x_1$ and a path  $p: I \rightarrow X$ such that $p(0) = x_0$ and $p(1)=x_1$, I want to show, If $\alpha \simeq_{x_1} \beta$ then 
$ \bar{p} * \alpha * p \simeq_{x_0} \bar{p} * \beta * p$. where $\bar{p}(t) = p(1-t)$ is a path with opposite direction.

My trial is using the definition of path product $\alpha* \beta(t)= \alpha(2t)$ for $0\leq t\leq \frac{1}{2}$ and $=\beta(2t-1)$ for $\frac{1}{2} \leq t \leq 1$. but i am confused with handling of $\bar{p} * \alpha * p(t)$. 
 A: I figure out what i was wrong. 
apparently 
\begin{align}
  \bar{p} * (\alpha * p)(t) = \begin{cases} \bar{p} (2t),  & 0 \leq t \leq \frac{1}{2} \\
\alpha * p (2t-1), & \frac{1}{2} \leq t\leq 1 \end{cases} 
= \begin{cases} p (1-2t),  & 0 \leq t \leq \frac{1}{2} \\
\alpha(4t-2)  & \frac{1}{2} \leq t \leq \frac{3}{4} \\
p(4t-3) & \frac{3}{4} \leq t \leq 1 \end{cases} 
\end{align}
and 
\begin{align}
  \bar{p} * (\beta * p)(t) = \begin{cases} \bar{p} (2t),  & 0 \leq t \leq \frac{1}{2} \\
\beta * p (2t-1), & \frac{1}{2} \leq t\leq 1 \end{cases}
= \begin{cases} p (1-2t),  & 0 \leq t \leq \frac{1}{2} \\
\beta(4t-2)  & \frac{1}{2} \leq t \leq \frac{3}{4} \\
p(4t-3) & \frac{3}{4} \leq t \leq 1 \end{cases}
\end{align}
we know a path is homotopy to itself, $p \simeq p$, the leftover parts is the region $\frac{1}{2} \leq t \leq \frac{3}{4}$. $i.e$, $\alpha(4t-2)$ and $\beta(4t-2)$ from our assumption $\alpha \simeq_{x_1} \beta$ it is guaranteed. Thus  $ \bar{p} * \alpha * p \simeq_{x_0} \bar{p} * \beta * p$
