Alternative derivation of $P(text{First qubit}=0)$ for the swap test

I’m trying to derive $P(text{First qubit}=0) = frac{1}{2} + frac{1}{2}|⟨a|b⟩|^2$ for the swap test.

The wiki page shows one way, but the result should also be obtainable via direct expansion of the tensor products, as attempted here. Present question is about bringing this process to completion as I — rather disappointingly — wasn’t able to finish the proof myself.

To summarize the thought process (adapted from here), let’s assume we start out with:

$$
|a rangle = a_0 |0rangle + a_1 |1rangle
|b rangle = b_0 |0rangle + b_1 |1rangle
|phi_0 rangle = |0 rangle |a rangle |b rangle
$$

Applying $H$ to $|phi_0rangle$, we get:

$$
|phi_1 rangle = H|0rangle|arangle|brangle = frac{1}{sqrt{2}}|0rangle|arangle|brangle + frac{1}{sqrt{2}}|1rangle|arangle|brangle
$$

Applying the swap, we get:

$$
|phi_2 rangle = frac{1}{sqrt{2}}|0rangle|arangle|brangle + frac{1}{sqrt{2}}|1rangle|brangle|arangle
$$

Applying the second H, we get:

$$
|phi_3 rangle = H|phi_2rangle =
frac{1}{2}|0rangle|arangle|brangle + frac{1}{2}|1rangle|arangle|brangle + frac{1}{2}|0rangle|brangle|arangle – frac{1}{2}|1rangle|brangle|arangle =
$$
$$
frac{1}{2}|0rangle left[|arangle|brangle + |brangle|arangleright] + frac{1}{2}|1rangle left[|arangle|brangle – |brangle|arangle right] tag{1}label{1}
$$

If we want to calculate $P(text{First qubit}=0)$, then the interesting quantity from $|phi_3rangle$ is the first term of eqref{1}:

$$
frac{1}{2}|0rangle left[|arangle|brangle + |brangle|arangleright]
=
frac{1}{2} left[
2 a_0 b_0 |0,0rangle
+ left( a_0 b_1 + a_1 b_0 right) |0,1rangle
+ left( a_0 b_1 + a_1 b_0 right) |1,0rangle
+ 2 a_1 b_1 |1,1rangle
right] =
$$
$$
a_0 b_0 |0,0rangle
+ frac{1}{2} left( a_0 b_1 + a_1 b_0 right) |0,1rangle
+ frac{1}{2} left( a_0 b_1 + a_1 b_0 right) |1,0rangle
+ a_1 b_1 |1,1rangle tag{2}label{2}
$$

Where I used:
$$
|a rangle |b rangle =
a_0 b_0 |0,0rangle
+ a_0 b_1 |0,1rangle
+ a_1 b_0 |1,0rangle
+ a_1 b_1 |1,1rangle

|b rangle |a rangle =
a_0 b_0 |0,0rangle
+ a_1 b_0 |0,1rangle
+ a_0 b_1 |1,0rangle
+ a_1 b_1 |1,1rangle
$$

Squaring the amplitudes from eqref{2}, we get:

$$
|a_0 b_0|^2
+ frac{1}{4} | a_0 b_1 + a_1 b_0 |^2
+ frac{1}{4} | a_0 b_1 + a_1 b_0 |^2
+ |a_1 b_1|^2 =
$$
$$
|a_0 b_0|^2
+ frac{1}{2} | a_0 b_1 + a_1 b_0 |^2
+ |a_1 b_1|^2 tag{3}label{3}
$$

I’m trying to prove that eqref{3} equals $frac{1}{2} + frac{1}{2}|⟨a|b⟩|^2$ but I could not.
I tried using:

EDIT: The following equations do not hold, as explained by C. Kang in the answers.

$$
|langle a| b rangle|^2 =
langle a| b ranglelangle a| b rangle^dagger =
langle a| b ranglelangle b| a rangle =
langle b| langle a| b rangle |arangle =
|a_0 b_0|^2 + |a_1 b_0|^2 + |a_0 b_1|^2 + |a_1 b_1|^2
$$
$$
|a_0 b_0|^2 + |a_1 b_1|^2 = |langle a| b rangle|^2 – |a_1 b_0|^2 – |a_0 b_1|^2 tag{4}label{4}
$$

Substituting eqref{4} into eqref{3}, I got:

$$
|langle a| b rangle|^2 – |a_1 b_0|^2 – |a_0 b_1|^2 + frac{1}{2} | a_0 b_1 + a_1 b_0 |^2
$$

But I’m not sure how to turn this into $frac{1}{2} + frac{1}{2}|⟨a|b⟩|^2$. Any thoughts?