I am trying to understand how the reset gate in Qiskit affects qubits its entangled with. Consider the following circuit with qubits $q_0$ and $q_1$:
Where circuit240 takes $|0\rangle$ to $a|0\rangle + b|1\rangle$ and circuit 244 takes $|0\rangle$ to $c|0\rangle + d|1\rangle$. Right before the reset gate on qubit $q_1$ the state of this circuit is $$\frac{1}{\sqrt{2}}(c|1\rangle + d|0\rangle)|0\rangle + \frac{1}{\sqrt{2}}(a|1\rangle + b|0\rangle)|1\rangle \tag{1}$$
I'm not quite sure how to mathematically represent what this quantum reset gate does to the quantum entangled state. For example, I tried a few tests with this circuit where I played with the values $a$ and $c$ and tested how the quantum reset gate affects the measurement of the qubit $q_0$. It seems that whether or not the quantum reset gate is added, it doesn't affect the measurements of the other entangled qubit. Does this generally hold?
Furthermore, when I take the qubit expression above and just reset qubit $q_1$to $|0\rangle$ I get the following:
$$\frac{1}{\sqrt{2}}(c|1\rangle + d|0\rangle)|0\rangle + \frac{1}{\sqrt{2}}(a|1\rangle + b|0\rangle)|0\rangle \tag{2}$$
$$= \frac{1}{\sqrt{2}}((a + c)|1\rangle + (b + d)|0\rangle)|0\rangle \tag{3}$$
But, mathematically, the probability of measuring $q_0$ as $|0\rangle$ in $(3)$ is not the same as the probability in qubit expression $(1)$(even though the tests show that removing the reset gate did not change the probability of measuring a $|0\rangle$ in the qubit $q_0$. What is the correct way to represent what the qubit reset gate does to an entangled qubit?