We can construct a 1/4 OT from 1/2 OT. Proving security for the receiver in 1/4 OT means showing a reduction to the 1/2 OT receiver's security. The protocol I am using is given on page 3 here. I am working with the simulation based definitions of security.
There are three 1/2 OT's used, meaning three reduction steps, all that will look very similar. I write the first explicitly below. R2 - real view of party one in 1/4 OT. R2' - the view of the first reduction step towards the simulator.
$$R2 \; ((m_0,m_1,m_2,m_3),(c_0,c_1)) = \lbrace(c_0,c_1), (s_0 \oplus s_2 \oplus m_0), (s_0 \oplus s_3 \oplus m_1), (s_1 \oplus s_4 \oplus m_2), (s_1 \oplus s_5 \oplus m_3), a, b, c \rbrace$$
where $s_0, s_1, s_2, s_3, s_4, s_5$ are all coin flips - the messages are booleans also. And $$\begin{aligned} a &\leftarrow R_21/2OT ((s_0, s_1), c_0),\\ b &\leftarrow R_21/2OT ((s_2, s_3), c_1),\\ c &\leftarrow R_21/2OT ((s_4, s_5), c_1)\end{aligned}$$ are samples from the real view of party 2 from the 1/2OT.
and
$$R2' \; ((m_0,m_1,m_2,m_3),(c_0,c_1)) = \lbrace(c_0,c_1), (s_0 \oplus s_2 \oplus m_0), (s_0 \oplus s_3 \oplus m_1), (s_1 \oplus s_4 \oplus m_2), (s_1 \oplus s_5 \oplus m_3), a, b, c \rbrace$$
where $s_0, s_1, s_2, s_3, s_4, s_5$ are all coin flips - the messages are booleans also. And $$\begin{aligned} a &\leftarrow S_21/2OT (c_0) ,\\ b &\leftarrow R_21/2OT ((s_2, s_3), c_1),\\ c &\leftarrow R_21/2OT ((s_4, s_5), c_1)\text{.}\end{aligned}$$
Let us assume that the above two distributions are distinguishable, by distinguisher $D$, we need to construct an adversary $A$ that distinguishes $R_21/2OT ((m, m'),c)$ and $S_21/2OT (c)$ for some $m,m',c$. $A$ receives as input one of the views it must distinguish and $c$ and outputs a boolean value.
In essence we must reconstruct the rest of the view (that is the same in R2 and R2') and pass it to $D$. The problem being the adversary does not know the messages $(m_0,m_1,m_2,m_3)$ and so cannot reconstruct the view totally faithfully.
I may be missing something obvious but if someone could explicitly construct the adversary for me to show this first step that would be great.
For avoidance of doubt. $A$ takes in as input either the distribution $R_21/2OT$ or $S_21/2OT$ and the $c$ (input) that was used to create these distributions and must output True or False depending on which distribution it was given. It can also access the distinguisher $D$ we have assumed can distinguish $R2$ and $R2'$.
