Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 1 - Section 1.8 - Proof Methods and Strategy - Supplementary Exercises - Page 112: 16

Answer

$S$ is not $Proposition$.

Work Step by Step

Let $p$= "$S$ is true" $q$= "$U$nicorns live" Using the above interpretation we can then rewrite S as” $(p \rightarrow q) \rightarrow q$” Let us first assume that $S$ is true, and then $p$ is true. However we cannot determine that $q$ is true or false, because the statement will be true if $q$ is true ($(p \rightarrow q) \rightarrow q$ is then false) and the statement will also be true if $q$ is false ($(p \rightarrow q) \rightarrow q$ is then true) Next, let us assume that $S$ is false, then$p \rightarrow q$ is true and $q$ is false. But then we can also know that $p$ is false (since $q$ is false and $p \rightarrow q$ is true). If $p$ is false, then this also states that $S$ is false. Thus we note that this assumption is also correct. Conclusion: we note that we cannot determine if $S$ is true of if $S$ is false (both options are possible). By the definition of a proposition, $S$ is then not a proposition.
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