6.7: Proof by counterexample
Sometimes we want to prove that \(P \not\Rightarrow Q\text{;}\) i.e. that \(P \rightarrow Q\) is not a tautology.
Recall. The equivalence
holds for any set of cases \(C_1, C_2, \ldots, C_m\) such that \(C_1 \lor \cdots \lor C_m\) is a tautology. (See Section 6.4 .)
So if \(P \land C_i \rightarrow Q\) is not a tautology for at least one \(i\text{,}\) then \(P \rightarrow Q\) also cannot be a tautology. Again, this also works in the universal case since \(\forall\) distributes over \(\land\) ( Proposition 4.2.1 ).
relative to the logical implication \(P \Rightarrow Q\text{,}\) a statement \(C\) such that \(P \land C \rightarrow Q\) is false
In Exercise 6.12.8, you are asked to prove the following statement by proving the contrapositive.
If \(2^n - 1\) prime, then \(n\) is prime.
Prove that the converse of this statement is false.
Solution
The converse statement is “If \(n\) is prime, then \(2^n - 1\) is prime.” But the case \(n=11\) is a counterexample :
is not prime even though \(n = 11\) is prime.
Check your understanding. Attempt Exercise 6.12.9 .