91Ó°ÊÓ

Give a counterexample to disprove each statement, where \(P(x)\) denotes an arbitrary predicate. Every prime number is odd.

Four women, one of whom was known to have committed a serious crime, made the following statements when questioned by the police: (B. Bissinger, Parade Magazine, 1993 ) $$\begin{array}{ll}{\text { Fawn: }} & {\text { "Kitty did it" }} \\ {\text { Kitty: }} & {\text { "Robin did it." }} \\ {\text { Bunny: }} & {\text { "I didn't do it" }} \\ {\text { Robin: }} & {\text { "Kitty lied." }}\end{array}$$ If exactly one of these statements is false, identify the guilty woman.

Write the converse, inverse, and contrapositive of each implication. If the calculator is working, then the battery is good.

Write the converse, inverse, and contrapositive of each implication. If London is in France, then Paris is in England.

Find the flaw in the following "proof": Let \(a\) and \(b\) be real numbers such that \(a=b .\) Then \(a b=b^{2}\) Therefore, \(a^{2}-a b=a^{2}-b^{2}\) Factoring, \(a(a-b)=(a+b)(a-b)\) Cancel \(a-b\) from both sides: $$a=a+b$$ since \(a=b,\) this yields \(a=2 a\) Cancel a from both sides. Then we get \(1=2\) . Let \(a, b,\) and \(c\) be any real numbers. Then \(a

Use De Morgan's laws to verify each. (Hint: \(p \rightarrow q \equiv \sim p \vee q\) ). $$\sim(p \rightarrow q) \equiv p \wedge \sim q$$

Let \(t\) be a tautology and \(p\) an arbitrary proposition. Find the truth value of each. $$(p \vee t) \rightarrow t$$

Let UD \(=\) set of integers, \(\mathrm{P}(x, y) : x\) is a multiple of \(y,\) and \(Q(x, y) : x \geq y\) Determine the truth value of each proposition. $$(\forall x) | P(x, 3) \rightarrow Q(x, 3) ]$$

Simplify each boolean expression. $$(p \wedge \sim q) \vee(p \wedge q) \vee r$$

Express \(p\) XOR \(q\) in terms of the Sheffer stroke. (Hint: \(\mathrm{XOR} q=[(p \vee q) \wedge \sim(p \wedge q)] .\)