Is this diagram wrong as the congruent circuit?
The second diagram is correct, because imposing \$V_2=0\$ means that no voltage drops across that portion of the circuit and such a condition is obtained indeed by short-circuiting \$r_2\$. This condition does not imply no current flow, and in fact \$I_2\$ flows through \$r_3\$, down into the shorted branch (by-passing \$r_2\$ and thus avoiding any voltage drop across it) and back towards the negative &quot;side&quot; of \$V_1\$, where it is combined with \$I_1-I_2\$ correctly yielding \$I_1\$.
Put it another way, if you actually opened your rightmost branch, then no current would flow through neither \$r_2\$ or \$r_3\$, but \$V_2=V_1\neq0\$ would hold and contradict your starting \$V_2=0\$ assumption.
Summarizing:

\$I=0\$ --&gt; open circuit
\$V=0\$ --&gt; short circuit