(d) i, iii, iv
iv) is expansion for natural join represented with other operators.
Why ii is not equivalent? Consider the following instances of R and S
$R: \left\{\left\langle``1", ``abc", ``p1", ``p2", ``p3"\right\rangle, \\ \left\langle``2", ``xyz", ``p1", ``p2", ``p3"\right\rangle \right\}$
$ S: \left\{\left\langle``1", ``abc", ``q1", ``q2", ``q3"\right\rangle \\ \left\langle``2", ``def", ``q1", ``q2", ``q3"\right\rangle \right\}$
Now, consider the given queries:
i. $R \bowtie S$ gives
$\left\{\left\langle``1", ``abc", ``p1", ``p2", ``p3", ``q1", ``q2", ``q3" \right\rangle \right\}$
Projecting $P$ gives $\left\{\left\langle ``1" \right\rangle \right\}$
ii. $\pi_P\left(R\right) \bowtie \pi_P\left(S\right)$ gives
$\left\{\left\langle``1" \right\rangle \left\langle ``2" \right\rangle \right\} \bowtie \left\{\left \langle ``1" \right\rangle \left\langle ``2"\right\rangle \right\}$
$=\left\{\left\langle ``1", ``2" \right\rangle \right\}$
iii. $\Pi_P \left(\Pi_{P, Q} \left(R\right) \cap \Pi_{P,Q} \left(S\right) \right)$ gives
$\left\{\left\langle``1", ``abc" \right\rangle, \left\langle ``2", ``xyz" \right\rangle \right\} \cap \left\{\left \langle ``1", ``abc" \right\rangle, \left\langle ``2", ``def" \right \rangle \right\}\\ = \left\{\left\langle``1", ``bc" \right\rangle \right\}$
Projecting $P$ gives $\left\{\left\langle ``1" \right\rangle \right\}$
iv. $\Pi_P \left(\Pi_{P, Q} \left(R\right) - \left(\Pi_{P,Q} \left(R\right) - \Pi_{P,Q} \left(S\right)\right)\right)$ gives
$\left\{\left\langle``1", ``abc" \right\rangle, \left\langle ``2", ``xyz" \right\rangle \right\} \\- \left( \left\{\left\langle``1", ``abc" \right\rangle, \left\langle ``2", ``xyz" \right\rangle \right\} - \left\{\left\langle``1", ``abc" \right\rangle, \left\langle ``2", ``def" \right\rangle \right\} \right) $
$= \left\{\left\langle``1", ``abc" \right\rangle, \left\langle ``2", ``xyz" \right\rangle \right\} \\- \left\{\left\langle ``2", ``xyz" \right\rangle \right\} \\ = \left\{\left\langle``1", ``abc" \right\rangle \right\}$
Projecting $P$ gives $\left\{\left\langle ``1" \right\rangle \right\}$