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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

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  • $\begingroup$ 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. $\endgroup$
    – paulinho
    Jul 28, 2020 at 15:17
  • $\begingroup$ Edited post, but wasn’t sure what to make the the letters ‘A’ after 1., 2., and 3. $\endgroup$ Jul 28, 2020 at 15:24
  • $\begingroup$ @ThomasAndrews I'm guessing "Assumption", though they are given as premises in the title. $\endgroup$
    – amWhy
    Jul 28, 2020 at 15:26
  • $\begingroup$ @amWhy Well premises are undischarged assumptions. $\endgroup$ Jul 28, 2020 at 23:19
  • $\begingroup$ @GrahamKemp Sure. Thanks for confirming my response to Thomas Andrews. $\endgroup$
    – amWhy
    Jul 28, 2020 at 23:21

2 Answers 2

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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.

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  • $\begingroup$ Justice is the lost word nowadays... $\endgroup$
    – Mikasa
    Jul 30, 2020 at 3:58
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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...

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