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everyone. I'm running into real trouble figuring out the following sentential logic problem.

  1. (S v T) ⊃ (S ⊃ ~T)
  2. (S ⊃ ~T) ⊃ (T ⊃ K)
  3. S v T

What we're supposed to get: S v K

Any help that anyone might offer on how to solve this one would be much appreciated. Thank you in advance.

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  • What is your question?
    – Natalie Clarius
    Aug 1, 2021 at 20:47
  • I'm stuck on how to use the rules for deductive reasoning to derive the conclusion "S v G" from the premises provided. I was hoping someone might be able to help me with this.
    – sixo33
    Aug 1, 2021 at 21:11
  • But which rules or steps are you stuck with specifically? What have you done that didn't work? As it is worded right now, this post contains no real question and reads like just a request for others to do your homework for you.
    – Natalie Clarius
    Aug 1, 2021 at 21:24

2 Answers 2

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General strategy: (1) If you're given a conditional and also given the antecedent of that conditional, just derive the consequent, because it simplifies the propositions available for you to use in the proof. (2) If you can't think of how best to proceed, try assuming the negation of the conclusion and deriving a contradiction.

  1. (S v T) => (S => ~T) [given premise]
  2. (S => ~T) => (T => K) [given premise]
  3. S v T [given premise]
  4. S => ~T [modus ponens from 1 and 3]
  5. T => K [modus ponens from 2 and 4]
  1. ~(S v K) [assumption]
  2. ~S ^ ~K [DeMorgan's Law from 6]
  3. ~S [conjunction elimination from 7]
  4. ~K [conjunction elimination from 7]
  5. ~T [modus tollens from 5 and 9]
  6. ~S ^ ~T [conjunction introduction from 8 and 10]
  7. ~(S v T) [DeMorgan's Law from 11]
  8. (S v T) ^ ~(S v T) [conjunction introduction from 3 and 12]
  1. ~~(S v K) [reductio ad absurdum from 6-13]
  2. S v K [double negation elimination from 14]
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  • Why are you using assumptions ro prove the conclusion? You do realize this can be proved with zero assumptions.
    – Logikal
    Aug 1, 2021 at 22:43
  • Yes, I realize that. Any tautology can be proven without assumptions. But assumption is a perfectly valid derivation strategy. Is there any special merit to derivations that abstain from the use of reductios? I also wasn't looking for the most elegant (least steps) proof; I was looking for the most useful strategy for someone practicing proofs. If I wanted to do it in fewer steps: 6. ~S=>T [Impl from 3]. 7. ~S=>K [HS from 5&6]. 8. ~~S v K [Impl from 7]. 9. S v K [DN from 8].
    – Dayv87
    Aug 1, 2021 at 23:39
  • By the use of the terminology I would say you learned so called logic from Math. Some terminology is the exact same words, pronunciation and spelling but the CONTEXT is absolutely not the same. Tautology has more than one context as do contradiction, contraposition, etc. I mentioned a proof with zero assumptions because math people tend to be trained THEY MUST USE ASSUMPTIONS. You seemed to so what you were trained to do in a math way. Many proofs -that are not as you call tautology- can have zero assumptions. In philosophy we usually do not use assumptions if the proof doesn't require them.
    – Logikal
    Aug 1, 2021 at 23:44
  • (1) I'm a Philosophy PhD with no math credentials. (2) I proved it without assumptions in my last comment. (3) A "tautology" is just a proposition that comes out true no matter what truth-values are assigned to its elements. For any proposition that can be proven in propositional logic, the conclusion in conjunction with the given premises must be a tautology for it to be provable. (4) Philosophers use assumptions all the time. In a reductio argument, you pose an assumption, derive a contradiction from it, and thereby prove the opposite of the original assumption.
    – Dayv87
    Aug 2, 2021 at 6:29
  • Thanks for your reply. The word tautology also refers to a pair of propositions when compared hold the same truth value. Tautology includes equivalent propositions & identical propositions using different words. You seem to be using the term tautology as math people do. You are saying one proposition is a tautology instead of it being a relationship of two or more propositions. Yes I am aware assumptions can be used. My point was math people believe there MUST BE an assumption. They typically begin their proofs with an assumption. In my philosophy classes we would use inference rules.
    – Logikal
    Aug 2, 2021 at 6:49
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The premises clearly allow deriving T ⊃ K through modus ponens, and with that you can readily derive S v K through a proof by cases. Here's the skelleton.

  1|  (S v T) ⊃ (S ⊃ ~T)    Premise
  2|  (S ⊃ ~T) ⊃ (T ⊃ K)    Premise
  3|_  S v T                 Premise
  4|  ......                 
  5|  T ⊃ K                 
  6|  |_ ......              
  7|  |  ......              
   |  +
  8|  |_ ......              
  9|  |  ......              
 10|  |  ......              
 11|  S v K
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