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Need help exercise using the FITCH program format. I'm stuck on where to start. The following 4 steps must be used to prove the goal.

P v Q
Q→ ¬ R
¬ P
¬ R → ¬ S

GOAL: ¬ S

Now I know: ¬ P and P v Q is true - hence, Q is true. Q is true and Q→ ¬ R is true. As true implies true statement, ¬ R is true. ¬ R → ¬ S is true. Since true never implies a false statement ¬ S is true. I just don't know where to start.

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  • Isn't that already the proof? As PvQ is true and as P is not true you know that Q must be true and as from Q follows not R you get not R from which follows not S qed. So the problem isn't were to start but how to write that in this weird system right?
    – haxor789
    Commented Jan 16 at 13:13
  • Could anyone please edit the formatting of the question to make it a bit more polished? Thank you
    – Julius Hamilton
    Commented Jan 16 at 14:53

2 Answers 2

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Now I know: ¬ P and P v Q is true - hence, Q is true. Q is true and Q→ ¬ R is true. As true implies true statement, ¬ R is true. ¬ R → ¬ S is true. Since true never implies a false statement ¬ S is true. I just don't know where to start.

That is basically it. The only question is whether you can use the rule of Disjunctive Syllogism, or need to build a proof by cases using fundamental rules of inference.

 a.| ¬P 
 b.| P v Q
   :
 c.| Q       DS a, b        | Disjunctive Syllogism
   :
   :
 d.|  |_ P   A              | Proof by Cases
 e.|  |  #   ¬E d,a         |
 f.|  |  Q   X  e           |
   :                        |
 g.|  |_ Q   A              |
   :                        |
 h.|  Q      vE b,d-f,g-g   |
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You will need to use Disjunction Elimination (Proof by cases). As P v Q is one of your premises, assume P and derive a sentence, then assume Q and derive that same sentence; you are now allowed to discharge it. The only remaining step is using Implication Elimination to reach the goal.

1. |  P v Q         
2. |  Q → ¬ R       
3. |  ¬ P           
4. |_ ¬ R → ¬ S     
5. |  |_ P          
6. |  |  ⊥          ⊥ Intro: 5,3
7. |  |  ¬ R        ⊥ Elim: 6
8. |  |_ Q          
9. |  | ¬R          → Elim: 2,8
10.|  ¬R            v Elim: 1, 5-7,8-9
11.| ¬ S            → Elim: 4,10
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  • This makes sense, but I'm unsure if my edits above are correct.
    – eaglefern
    Commented May 14, 2020 at 1:49
  • If you meant Negation Elimination (where you wrote INTRO), then that step is correct. I do not understand the step where you annotated elim (6). Can you clarify ?
    – F. Zer
    Commented May 14, 2020 at 2:17
  • elim ⊥ is eliminate in reference to step 6 (the step before). And the reit is reiterating premise 2
    – eaglefern
    Commented May 14, 2020 at 2:20
  • In step 6, you are deriving P v Q, but that is one of your premises. Can you use Explosion rule ? Also, no need to reiterate premise 2, I think. Does your system require that step ?
    – F. Zer
    Commented May 14, 2020 at 2:32
  • No, I can't use Explosion rule. Disjunction Elimination is certainly correct, but I'm not sure how to add it on Fitch
    – eaglefern
    Commented May 14, 2020 at 2:44

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