I'm having a difficult time deriving ~(RvB) from ~R and ~B. This would be fairly straight-forward if we were allowed to use the DeMorgan Laws, however my professor is not allowing us to use them for this problem set. Any help would be much appreciated.

The rules that I am allowed to use are:

  • Conjunction Elimination/Introduction
  • Conditional Elimination/Introduction
  • Negation Elimination/Introduction
  • Disjunction Elimination/Introduction
  • Biconditional Elimination/Introduction


closed as too broad by Joseph Weissman Jan 9 '16 at 16:22

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  • If you're familiar with DeMorgan's Laws, you're presumably familiar with their derivation. That's all you need. – WillO Feb 22 '15 at 14:19

Assume RvB for use with Negation Introduction, and then use disjunction elimination to work with ~R and RvB to get B: RvB says that at least one of those two sentences, R and B, must be true, and ~R says that R isn't true. It follows that B is true. But ~B. So, a contradiction results from assuming RvB, and you can conclude ~(RvB).


A strategy that tends to result in short and easy to understand proofs is to immediately assume the negation of the sentence you're trying to prove and then make derivations until you find a contradiction. Having found a contradiction allows you to deduce the negation of the sentence you assumed that lead to the contradiction (this is a valid inference rule called reductio ad absurdum). Here is an example:

  1. ~R
  2. ~B
  3. |-asm: (R v B)
  4. |-∴ B { from 1 and 3 }
  5. ∴ ~(R v B) { from 3; 2 contradicts 4 }

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