I'm trying to prove an argument of the form:

  1. B
  2. ~(C & B)

Therefore: ~C.

I can expand out ~(C & B) into ~C OR ~B, and with the premise B, it is clear that ~C is the case.

However, I'm having trouble proving this using a Fitch style system. I've tried disjunctive elimination, but I can't see how to get to ~C from an assumption of ~B, so I'm wondering whether or not this is the correct way to go.

If anyone knows how to show the equivalent of a disjunctive syllogism in Fitch, or at least somewhere to find out how, some direction would be greatly appreciated.

  • 2
    Assume C. Conjoin C and B to get (C & B), which contradicts premise 2. Conclude ~C. [the only rules used here are: &-Intro, Bottom/Absurd-Intro, ~-Elim]. – Hunan Rostomyan Oct 10 '13 at 22:50
  • Indeed, thank you. I just figured that out. I was making it much harder than it was by focusing on disjunctive elimination. – Sinthet Oct 10 '13 at 22:53
  • Here's one with De Morgan & v-Elim: Push the negation in premise (2) to get (~C v ~B). Assume ~C. Reiterate to get ~C. Assume ~B to get a contradiction with premise (1) and conclude ~C. Since ~C follows from both cases, it follows by v-Elim that ~C. – Hunan Rostomyan Oct 10 '13 at 22:57
  • 1
    +1 Shows research effort, is clear, and might even be useful to others. No less importantly, it has the appropriate tags. – Hunan Rostomyan Oct 11 '13 at 0:11

Here are a couple options:

  • If you have a double negation rule, you can turn B into ~~B. Then you can use a disjunctive syllogism rule together with (~C v ~B) to get ~C.

  • You can try an indirect proof, where you assume C, and then conjoin it with B to get (C & B), which yields a contradiction with line 2, entailing ~C.

  • Hey, thanks for replying! The problem I'm having is due to the absence of a disjunctive syllogism rule. The only way to do an v elimination in Fitch is via Proof by Cases (where the cases are sub-proofs). If I go with an indirect proof and get (C & B), then all I can do within Fitch is introduce a contradiction, from which I can derive ~(C & D), but then I'm back where I started again. An overview of the system Im working in. – Sinthet Oct 10 '13 at 22:42
  • And.... Disregard that :). I indeed solved it with an indirect proof, not quite sure how I overlooked that possibility. I was so focused on using DeMorgan's Laws in my proof that I forced myself into thinking disjunctive elimination had to be the way to proceed. Thank you! – Sinthet Oct 10 '13 at 22:52

Here is an approach that uses disjunctive elimination. Given the following I assume the DeMorgan rules are available:

I can expand out ~(C & B) into ~C OR ~B

enter image description here

On line 3, I used DeMorgan rules (DeM) to get a disjunction which I will then need to eliminate to reach the goal.

To eliminate the disjunction, I have to consider both disjuncts, "¬C" and "¬B".

The first disjunct is the easiest, but it may be confusing because it is so easy. The assumption, "¬C", for that subproof is precisely what I want to show. There is nothing more to do (at least for this proof checker).

The second disjunct uses explosion based on the observation that lines 1 and 5 are contradictory. This proof checker allows me to state the contradiction (⊥) on line 6 and use explosion (X) on line 7 to reach the conclusion that I want "¬C". The one you are using may require something different.

Since I have the same conclusion for each disjunct I can discharge the two assumptions on lines 4 and 5 by citing the rule of disjunction elimination (∨E). In this proof checker I have to reference the disjunction itself (3), the first subproof (which is only line 4 but written as a range 4-4), and the second subproof (5-7).


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, Winter 2018. http://forallx.openlogicproject.org/

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