How to prove $(p \lor q); (p\rightarrow r); (q\rightarrow s);$ therefore $(r \lor s),$ without modes tollens or derived rules for natural deduction

Question

How to prove

$(p \lor q);$

$(p\rightarrow r);$

$(q\rightarrow s);$

therefore $(r \lor s).$

without modes tollens or derived rules for natural deduction

I can prove the above set with derived rules, (negation rules, modus tollens, etc), but is there a way to prove it using only the basic natural deduction rules. Thank you

Answer 1

From the first premise, we have $p\lor q$.

$\qquad$ Assume p. Then r. (Modus ponens, first premise plus premise 2). Then $r\lor s$.

$\qquad$ Assume q. Then s. (Modus ponens, first premise, plus premise 3). Then $r\lor s$.

Therefore $r\lor s$.

I'll let you complete the justifications.

Answer 2

Okay, one of the premises is a disjunction, so a disjunction elimination would seem promising.

At the other end, the target is a disjunction, so disjunction introduction is also indicated.

Now, for the in-between, the other two premises are conditional statements, whose antecedents and consequents look most useful, so...