Problem 1
Given the following propositions generated by \(p, q,\) and \(r,\) which are equivalent to one another? (a) \((p \wedge r) \vee q\) (e) \((p \vee q) \wedge(r \vee q)\) (b) \(p \vee(r \vee q)\) (f) \(r \rightarrow p\) (c) \(r \wedge p\) (g) \(r \vee \neg p\) (d) \(\neg r \vee p\) (h) \(p \rightarrow r\)
Problem 1
Prove with truth tables: (a) \(p \vee q, \neg q \Rightarrow p\) (b) \(p \rightarrow q . \neg q \Rightarrow \neg p\)
Problem 3
Give direct and indirect proofs of: (a) \(a \rightarrow b, c \rightarrow b, d \rightarrow(a \vee c), d \Rightarrow b\). (b) \((p \rightarrow q) \wedge(r \rightarrow s),(q \rightarrow t) \wedge(s \rightarrow u), \neg(t \wedge u), p \rightarrow r \Rightarrow \neg p .\) (c) \(p \rightarrow(q \rightarrow r), \neg s \vee p, q \Rightarrow s \rightarrow r .\) (d) \(p \rightarrow q_{1} q \rightarrow r_{1} \neg(p \wedge r), p \vee r \Rightarrow r\) (e) \(\neg q \cdot p \rightarrow q \cdot p \vee t \Rightarrow t\)
Problem 3
Over the universe of booles, define the propositions \(B(x): x\) has a blue cover, \(M(x): x\) is a mathematics book, \(U(x): x\) is published in the United States, and \(R(x, y):\) The bibliography of \(x\) includes \(y\). Translate into words: (a) \((\exists x)(\neg B(x))\). (b) \((\forall x)(M(x) \wedge U(x) \rightarrow B(x))\) (c) \((\exists x)(M(x) \wedge \neg B(x))\) (d) \((\exists y)((\forall x)(M(x) \rightarrow R(x, y))) .\) (e) Express using quantifiers: Every book with a blue cover is a mathematics book. (f) Express using quantifiers: There are mathematics books that are published outside the United States. (g) Express using quantifiers: Not all books have bibliographies.
Problem 4
Prove that for \(n \geq 0: \sum_{k=0}^{n} 2^{k}=2^{n+1}-1\).
Problem 6
Use the following definition of absolute value to prove the given statements: If \(x\) is a real number, then the absolute value of \(x_{1}|x|\), is defined by: $$ |x|=\left\\{\begin{array}{cl} x & \text { if } x \geq 0 \\ -x & \text { if } x<0 \end{array}\right. $$ (a) For any real number \(x,|x| \geq 0 .\) Moreover. \(|x|=0\) implies \(x=0\). (b) For any two real mumbers \(x\) and \(y,|x| \cdot|y|=|x y|\). (c) For any two real numbers \(x\) and \(y,|x+y| \leq|x|+|y|\).