# 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

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

• If $p \vee q$ is true, then at least one of $p$ or $q$ is true. You can consider the individual cases and calculate $r \vee s$ in each case, showing that it is always true. Jul 28, 2020 at 15:17
• Edited post, but wasn’t sure what to make the the letters ‘A’ after 1., 2., and 3. Jul 28, 2020 at 15:24
• @ThomasAndrews I'm guessing "Assumption", though they are given as premises in the title. Jul 28, 2020 at 15:26
• @amWhy Well premises are undischarged assumptions. Jul 28, 2020 at 23:19
• @GrahamKemp Sure. Thanks for confirming my response to Thomas Andrews. Jul 28, 2020 at 23:21

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.

• Justice is the lost word nowadays... Jul 30, 2020 at 3:58

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...