Here are the premises/conclusion, and where I've gotten so far.

$1.$ $(W\wedge E)\rightarrow (P\vee L)$ (PR)

$2.$ $(W\wedge \neg E)\wedge R))\rightarrow (P\vee D)$ (PR)

$3.$ $((W\wedge \neg E)\wedge \neg R))\rightarrow (U\wedge \neg F)$ (PR)

$-.$ $(W\wedge F)\rightarrow (L\vee D)\vee P$ (CO)

$4.$ $\neg(U\wedge \neg F)\rightarrow \neg(W\wedge(\neg E\wedge \neg R))$ From $3$ transposition

$5.$ $(\neg U\vee \neg \neg F)\rightarrow \neg(W\wedge(\neg E\wedge \neg R))$ From $4$ DeM

$6.$ $(\neg U\vee F)\rightarrow \neg(W\wedge(\neg E\wedge \neg R))$ From $5$ Double negation

$7.$ $(\neg U\vee F) \rightarrow (\neg W\vee \neg (\neg E\wedge \neg R))$ From $6$ DeM

$8.$ $(\neg U \vee F)\rightarrow(\neg W\vee(\neg \neg E\vee \neg \neg R))$ From $7$ DeM

$9.$ $(\neg U \vee F)\rightarrow (\neg W \vee (E\vee R))$ From $8$ double negation

Here is where my conditional proof starts.

$10.$ $W\wedge F$ (Provisional assumption)

$11.$ $W$ From $10$ conjunction elimination (CE)

$12.$ $F$ From $10$ CE

$13.$ $\neg U \vee F$ From $12$ disjunctive addition (DA)

$14.$ $(\neg W\vee (E\vee R))$ From $9$, $13$ Modus Ponens

$15.$ $E\vee R$ From $14$, $11$ Disjunctive syllogism

At this point I'm lost. I've tried assuming several negations for reductio (contradiction) but I haven't gotten the desired result. Any ideas?

  • $\begingroup$ I'm lost too. What rules of inference can you use? Do you have any axioms or axiom schema at hand? $\endgroup$ – Doug Spoonwood Oct 13 '14 at 6:17
  • $\begingroup$ Conjunction elimination, conjunction introduction, hypothetical syllogism, disjunctive addition/syllogism, disjunctive dilemma, modus tollens/ponens. It's possible, although unlikely, that I translated the argument incorrectly. I've triple checked it $\endgroup$ – user1227849937 Oct 13 '14 at 6:23

Up to step 15 it's Ok.

We can use the Law of Excluded Middle : $E \lor \lnot E$ and proceed by cases (Disjunction Elimination).


Assuming $E$, we have $W \land E$ by Conjunction Introduction with step 11.

Thus, from premise (1) we conclude with $(P \lor L)$ by Modus Ponens and then with $(P \lor (L \lor D))$ by Disjunction Introduction (re-arranging the terms).


Assuming $\lnot E$, we have $R$ from step 15 : $E \lor R$, by Disjunctive Syllogism. Thus we have $((W \land \lnot E) \land R)$ and we conclude from premise (2) by MP with : $(P \lor D)$, and again with $(P \lor (L \lor D))$ by Disjunction Introduction (re-arranging the terms).

Thus, from the assumption $(W \land F)$, we have proved $(P \lor (L \lor D))$ twice, both under assumption $E$ and under assumption $\lnot E$.

I.e., we have, by Conditional Proof :

$(W \land F) \vdash E \rightarrow (P \lor (L \lor D))$


$(W \land F) \vdash \lnot E \rightarrow (P \lor (L \lor D))$.

With the Law of Excluded Middle, by Disjunction elimination, we can conclude with :

$(W \land F) \vdash (P \lor (L \lor D))$.

and by Conditional Proof with :

$(W \land F) \rightarrow (P \lor (L \lor D))$.


If we nedd a proof of the tautology $E \lor \lnot E$, we can insert this "sub-proof" :

$W$ --- step 11 of the main proof

$E$ --- assumed for CP

$E \lor E$ --- by Disjunction introduction

$E$ --- by Tautology

$E \rightarrow E$ --- by CP

$\lnot E \lor E$ --- by Material implication.


The following proof using a Fitch-style proof checker attempts the problem directly by assuming the antecedent of the goal and deriving the consequent.

enter image description here

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.