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Section 6.7 Proof by counterexample
Sometimes we want to prove that
\(P \not\lgcimplies Q \text{;}\) that is, that
\(P \lgccond Q \) is
not a tautology.
Recall.
The equivalence
\begin{equation*}
P \lgccond Q \lgcequiv (P \lgcand C_1 \lgccond Q) \lgcand \dotsb \lgcand (P \lgcand C_m \lgccond Q)
\end{equation*}
holds for any set of cases
\(C_1, C_2, \dotsc, C_m \) such that
\(C_1 \lgcor \dotsb \lgcor C_m \) is a tautology. (See
Section 6.4 .)
So if
\(P \lgcand C_i \lgccond Q \) is
not a tautology for at least one
\(i \text{,}\) then
\(P \lgccond Q \) also cannot be a tautology. Again, this also works in the universal case since
\(\forall \) distributes over
\(\lgcand \) (
Proposition 4.2.7 ).
counterexample
relative to the logical implication
\(P \lgcimplies Q \text{,}\) a statement
\(C \) such that
\(P \lgcand C \lgccond Q \) is false
Worked Example 6.7.1 .
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 :
\begin{equation*}
2^{11} - 1 = 2047 = 23 \cdot 89
\end{equation*}
is not prime even though \(n = 11 \) is prime.
Check your understanding.
