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Exercise 3.23a of Kaufman, 2005: Find the Norton equivalent of the circuit in Figure 3.125: Figure 3.125 circuit

I've run into some trouble when answering this question. The Norton equivalent resistance is clearly \$R_N=R_6+R_7\$. The Norton equivalent current can be found using superposition.

In the first sub-circuit, \$I\$ is changed to an open circuit, and \$I_N=v/(R_6+R_7)\$. According to the answer I was provided, this is the only term. This implies that the current from the other sub-circuit, in which \$v\$ is changed to a short circuit, is \$0\$.

I don't see how this can be the case. My thinking is that from \$I\$, I need to use the current divider relation to determine what portion of the current passes through \$R_4\$, then again at the junction of the short circuit when \$v\$ used to be, and once again to see what portion passes through \$R_6\$.

If \$R_1\$ or \$R_2\$ were not present, the given answer would make sense to me, because all no current would flow through the resistors to the right of where \$v\$ was, but since they do exist, there is some resistance in all paths back to \$I\$, so the current divider formula should apply.

Can someone correct/affirm my thinking on this?

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v is a fixed voltage source, nothing connected to it changes the voltage, so you can ignore the entire circuit left of v as well as R5.

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