= Two-current model with unequal magnetic layers
{title2=$R_{AP}-R_P=(a_1-b_1)(a_2-b_2)/\Sigma$}
Let the first layer have channel resistances $a_1,b_1>0$ and the second $a_2,b_2>0$, with a common spacer $r\geq0$. Define $\Sigma=a_1+b_1+a_2+b_2+2r$. The same circuit rules give
$$
R_P=\frac{(a_1+a_2+r)(b_1+b_2+r)}{\Sigma},\qquad
R_{AP}=\frac{(a_1+b_2+r)(b_1+a_2+r)}{\Sigma}.
$$
Subtracting their numerators gives $(a_1-b_1)(a_2-b_2)$. The sign therefore depends on whether the two layers favor the same global spin channel. With $(a_1,b_1)=(1,4)\,\Omega$, $(a_2,b_2)=(4,1)\,\Omega$ and $r=1\,\Omega$, the values reverse: $R_P=3.00\,\Omega$, $R_{AP}=2.25\,\Omega$. The model's ratio $m$ is then $-1/4$. Positive magnetoresistance was a consequence of the identical-layer assumptions, not a theorem about every two-channel network.
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