= Solution
Retain $E:y^2=x(x^2+ax+b)$ and put $b'=a^2-4b$. The explicit <two-isogeny formula> is
$$
\phi(x,y)=\left(x+a+\frac bx,\ y\left(1-\frac b{x^2}\right)\right),\qquad
\psi(X,Y)=\left(\frac{X-2a+b'/X}{4},\ \frac{Y(1-b'/X^2)}8\right),
$$
with the maps extended at the kernel points. Direct substitution verifies the equations and $\psi\phi=[2]$.
Here is a concrete kernel test for the descent homomorphisms. A preimage of $(X,Y)\in E'$ has $x$ satisfying $x^2+(a-X)x+b=0$, whose discriminant is $(X-a)^2-4b$. The target equation says $Y^2=X((X-a)^2-4b)$, so for nonexceptional points this quadratic has a rational solution precisely when $X$ is a square. The resulting $y$ follows from the displayed <elliptic isogeny> formula. For a nonzero target two-torsion point, the preimage quadratic has a double root $x=(X-a)/2$, and the source equation gives $y^2=x^2X$, so the same square condition still applies. At $(0,0)\in E'$, a rational preimage exists precisely when $b'$ is a square, matching the special square-class value $\alpha'((0,0))=[b']$. The dual argument gives the other kernel. Thus
$$
A:=\alpha(E(\mathbb Q))\cong E(\mathbb Q)/\psi E'(\mathbb Q),\qquad
B:=\alpha'(E'(\mathbb Q))\cong E'(\mathbb Q)/\phi E(\mathbb Q).
$$
Let $\delta=[\ker\psi:\ker\psi\cap\phi E(\mathbb Q)]$. The quotient map induced by $\psi$ has this kernel, giving
$$
[E(\mathbb Q):2E(\mathbb Q)]=\frac{\#A\,\#B}{\delta}.
$$
If $b'$ is a rational square, $E$ has four rational two-torsion points and $\delta=1$; otherwise it has two and $\delta=2$. In either case $\delta\,\#E(\mathbb Q)[2]=4$. Combining with finite generation therefore proves the <two-isogeny rank formula>
$$
\boxed{2^{g_E}=\frac{\#A\,\#B}{4}.}
$$
This derivation checks the factor four, including curves with full rational two-torsion.
To compute these images, enumerate the signed square-free divisors $d$ of $b$ and $b'$. Substituting $x=d(U/V)^2$ yields the <quartic covering in a two-isogeny descent>
$$
W^2=dU^4+aU^2V^2+(b/d)V^4,\qquad(U,V)\ne(0,0).
$$
The class $d$ occurs precisely when this covering has a <rational point>, with the identity and two-torsion handled by the boundary cases. Test at the real place and at primes dividing $2bb'$ to rule out candidates, and search for <rational points> to prove that surviving classes occur. The locally soluble classes form a finite <elliptic isogeny> <Selmer group of an elliptic curve>, giving an upper rank bound. <rational points> and independent height pairings give a lower bound. When the bounds agree the rank is proved. \b[Local solubility alone does not prove a <rational point> exists]: if a gap remains, a higher descent or further information about the covering obstructions is needed. This is why the procedure usually computes the rank but is not an unjustified universal termination claim.
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