You've already answered part of your question yourself: Since the truth tables are the same, the expressions are equivalent, so it's not a glitch in the software.
To see that $(2)$ is equivalent to $(3)$, note that for $zx$ to have an effect, both $x$ and $z$ would have to be true; but in that case $xy$ is $y$ and $zy'$ is $y'$, and $y+y'$ is true; thus $zx$ is redundant. More formally:
$$
\begin{align}
xy+zy'+zx&=xy+zy'+zx(xy+zy')'
\\
&=xy+zy'+zx(xy)'(zy')'
\\
&=xy+zy'+zx(x'+y')(z'+y)
\\
&=xy+zy'+zxy'y
\\
&=xy+zy'
\end{align}$$
(where I moved $zx$ to the right to make it typographically apparent that the rest stays the same).
If the hypothetical syllogism is a theorem, then:
$$(p\overline{q})+(q\overline{r}) + \overline{p} + r = 1.$$
Here is one way of demonstrating that:
$$\overline{(p\overline{q})+(q\overline{r}) + \overline{p} + r} = 0.$$
$$(\overline{p} + q)(\overline{q} + r) p \overline{r} = 0.$$
$$p (\overline{p} + q)\overline{r}(\overline{q} + r) = 0.$$
$$(p\overline{p} + pq)(\overline{r}\overline{q} + \overline{r}r) = 0.$$
$$(0 + pq)(\overline{r}\overline{q} + 0) = 0.$$
$$(pq)(\overline{r}\overline{q}) = 0.$$
$$(p\overline{r})(q\overline{q}) = 0.$$
$$(p\overline{r})(0) = 0.$$
$$0 = 0.$$
I didn't read the comments, so I suspect you already got a satisfactory answer, but if not, this might be of some help. If some step is wrong or unclear, leave a comment and we'll find the appropriate rule of boolean algebra that justifies it.
Since you explicitly asked for a direct proof and I am stuck at an airport, I'll add the following:
$$(p\overline{q})+(q\overline{r}) + \overline{p} + r = 1.$$
$$(~(p\overline{q})+ \overline{p}~) + (~(q\overline{r}) + r~) = 1.$$
$$(~\overline{q} + \overline{p}~) + (~q + r~) = 1.$$
$$(~\overline{q} + q~) + (~\overline{p} + r~) = 1.$$
$$(1) + (~\overline{p} + r~) = 1.$$
$$1 + \dots = 1.$$
Best Answer
As far as I understand it correctly and $+$ means "or"; $\cdot$ means "and"; $\bar{x}$ means negation, it can surely be reduced. Btw, there is a mistake, because $(1)$ is not true if $x\bar y\bar z=1$, i.e., $(x,y,z)=(1,0,0)$, whereas $(2)$ obviously contains all $8$ cases and is therefore always true.
A general method to reduce such expressions is to "redistribute" by the following equivalent operations:
Using these we reduce $(1)$ to $\overline{x\bar y \bar z}$. There're surely algorithms for this (I don't know them), but it is similar to simplification of any formula, it takes some practice.