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This one is driving me crazy. I don't understand most keys for de morgan, modus ponens, etc, so please abbreviate if possible? EX: DM, MP, SIMP, HS, Conj, Imp (material Implication). Thank you anybody who can help out! v or, ~ not, ^ and, > if...then, => if and only if (bi-conditional).

  1. P>(RvS)
  2. ~[(~Pv~Q)v(Rv~L)]

and the conclusion is S.

So:

  1. ~(~Pv~Q)v~(Rv~L) 2,DM

  2. (PvQ)v(~RvL) 3,DM

  3. ~(PvQ)>(~RvL 4,IMP

  4. ~Pv(RvS) 1, IMP

  5. (~PvR)vS 6, ASSOC

  6. (Rv~P)vS

  7. ~Rv~P)>S

  8. (~RvP)>S

What's getting me is I can't isolate a variable that helps complete the conclusion that S is so! I have other notes, too, but can't place it. There is no other info than the first 2 lines. I can't isolate a truth because they're all disjunctions (v) or conditionals (>). Am I missing an Addition, Modus Tolens, or Modus Ponens somewhere? I was thinking maybe a double negation, but I can't figure this out after working on it almost all day yesterday. Final question on a Logic exam already passed. Hope this makes sense?

  • You're using De Morgan incorrectly. The disjunction should change to conjunction. – Eliran Jul 27 at 22:56
  • Think i got it solved – Mick Jul 28 at 0:34
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Here is a solution to compare with what you have. Also you might find the proof checker helpful to check the other proofs you are asked to do:

enter image description here

For this proof checker DeM is De Morgan rule, ∧E is conjunction elimination, DNE is double negative elimination, →E is conditional elimination and DS is disjunctive syllogism.


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

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Fall 2019. http://forallx.openlogicproject.org/forallxyyc.pdf

  • So you have to build the arguement yourself but it checks it for you? Am I missing something where you put in the 2 first premises and conclusion and it solves it for you? – Mick Jul 28 at 1:29
  • @Mick Yes, it only checks each step. You put in the two premises and the conclusion and then click "Create Problem". Each of the steps after that you have to enter on your own, but you can check each step by clicking the "Check Proof" button. – Frank Hubeny Jul 28 at 4:52
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|  1. P > (S v R)
|_ 2. ~((~P v ~Q) v (R v ~L)]  

To conclude S from the first premise you need to derive P and ~R from the second. So, judging by your rule abbreviations , your proof should look somewhat like>

|  3. ~(~P v ~Q) ^ ~(R v ~L)   DM 2       De Morgan's
|  4. ~(~P v ~Q)               SIMP 3     Simplification
|  5. ~~P ^ ~~Q                DM 4
|  6. ~~P                      SIMP 5
|  7. P                        DNE 6      Double Negation Elimination
|  8. :                        :
|  9. :                        :         ---similarly
| 10. ~R                       :
| 11. S v R                    MP 7, 1    Modus Ponens
| 12. S                        DS 11, 10  Disjunctive Syllogism

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