Tag Archives: reductio

Reductio Challenge

In formal proofs of validity, the reductio ad absurdum method can be used to make some proofs easier, and even some shorter. For example, consider this argument:

(~P ⊃ R) • (~Q ⊃ S)    ~(R S)    ∴ P • Q

The proof for this valid argument is 14 steps without the reductio (which I will let you try to solve on your own), but only 7 steps with the reductio, as shown here:

  1. (~P ⊃ R) • (~Q ⊃ S)
  2. ~(R ∨ S)   /  ∴  P • Q
  3. ~(P • Q)                     R.A.A.
  4. ~P ∨ ~Q                    3 De M.
  5. R ∨ S                         1, 4 C.D.
  6. (R ∨ S) • ~(R ∨ S)   5, 2 Conj.
  7. P • Q                          3-6 R.A.

The reasoning behind the reductio method is this: If assuming that a proposition is false leads to a self-contradiction, then the proposition must be true. This reasoning can itself be written as a propositional argument:

~P ⊃ (Q • ~Q)   ∴  P

This is a valid argument, as a shorter truth table will show. But the proof for this argument (if you are not allowed to use reductio) requires 13 steps, and it is rather difficult to solve. Any takers?