Why does the converse of the clock condition imply that any two concurrent events occur at the same time?
Assume the clock condition AND its converse: $$C(a)\lt C(b) \Rightarrow a \rightarrow b \tag{*}\label{*}$$
Next, take note that $\neg(a \rightarrow b)$ implies either $b \rightarrow a$ or $a \nrightarrow b$ (i.e. if '$a$ does NOT happen before $b$' then either '$b$ happens before $a$' or they are concurrent events).
Now, since the implication $\eqref{*}$ is (assumed) true, its contrapositive must also be true:
$$\neg(a \rightarrow b) \Rightarrow C(a)\geq C(b)$$
Applying our negation rule, we find either $b \rightarrow a \Rightarrow C(a)\geq C(b)$ (which reduces to the clock condition) or $$a \nrightarrow b \Rightarrow C(a)\geq C(b)$$
Lamport is only interested in the latter case: &quot;...we cannot expect the converse condition to
hold as well, since that would imply that any two $\mathbf{concurrent}$ events must occur at the same time.&quot;
To show this, remember Lamport said when $a,b$ are concurrent then both $a \nrightarrow b$ and $b \nrightarrow a$ so $$b \nrightarrow a \Rightarrow C(b)\geq C(a)$$
The last two equations imply if $a$ and $b$ are concurrent, $C(a)=C(b)$.
To derive the contradiction, take any two events $a, b$ concurrent but on separate processes. Then their clock numbers must be the same. Now if there is another concurrent event before or after $a$ (on the same process), then it too must have the same clock number as $b$ and therefore $a$. But that contradicts the clock condition (e.g. see condition C1 in the paper).