SQL, optimizing a select statement in relational algebra
I suspect this is a homework question.

Edit: OK, you're trying to invent your own homework. Well done for trying. You might have bitten off too much.

We need first to sort out the headings of your relations. As per my comment above, Cartesian product must have no attributes in common between operands. I'll assume headings: R(a,b), S(c,d), T(e,f) -- that is, no b in common between R, S; no d in common between S, T. Then I don't need to put the dot-prefix on attribute names. Your question becomes:

$\pi_{a,d} (\sigma_{(b = "d" \land d \geq 100) \lor f \geq 200}((R\times S)\times T))$

(If you do want attributes in common, then instead of $\times$, use Natural Join $\bowtie$. And the whole point about Natural Join (a beautiful operation, at the heart of the Relational Model) is that it 'commons up' the attribute names. So you still don't need the dot-prefix.

And what it's trying to teach you is that the Boolean operators (AND /\, OR \/) within a Select expression (sigma) are equivalent to certain set operators over the relations.

Edit: OK you've shown one equivalent for $\land$, which is to push the $\sigma$ inside the $\times$. More usually, you'd use set intersection $\cap$.

Yes you are allowed to 'split' a selection that is a disjunction (your sigma x\/y). Think about what is the set-theoretic operator that's equivalent to disjunction.

Edit: The set-theoretic operator is union $\cup$. For both intersection and union, we need the headings of the operands to be the same -- that is, the full Cartesian product.

Also beware: what's the relative precedence of the disjunction vs the conjunction in that sigma expression?

Edit: OK, you've answered that. So the outermost operator is $\lor$. Let's rewrite that to $\cup$ between relation values:

$\pi_{a,d} (\sigma_{(b = "d" \land d \geq 100)}((R\times S)\times T) \cup \sigma_{(f \geq 200)}((R\times S)\times T))$

Now in the left operand of $\cup$ we have a $\sigma$ with a $\land$. We can push that inside, as you've already shown. Similarly in the right operand we can push the $\sigma$ inside:

$\pi_{a,d} ( ((\sigma_{(b = "d")}R\times \sigma_{(d \geq 100)}S)\times T) \cup ((R\times S)\times \sigma_{(f \geq 200)}T))$

But notice in the right operand of $\cup$, we still have bare $(R \times S)$. So the union is going to include every (a,b) pair from S, and every (c,d) pair from R. (I suggest you try that out with a few sample sets of attribute values for those relations.) Your outermost projection, just takes a from R and d from S. So all the business with the $\sigma$s and $\land$, $\lor$ is rather pointless. And relation T is superfluous.

The most reduced form of your formula is:

$\pi_{a,d}(R \times S)$

Or even:

$(\pi_{a}(R)) \times (\pi_{d}(S))$