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F & F & F & T & T & T & T & T\\   F & F & T & T & T & T & T & T\\   \end{tabular}   \end{table} \end{table}\\  12) (e$\wedge$p)$\rightarrow$m\\  16)\\  a) q$\ringtarrow$p\\  b) q$\wedge$$\neg$p\\  c) q$\rightarrow$p\\  d) $\neg$q$\rightarrow$$\neg$p\\  \section(Chapter 2 solution:)\\  4)\\  a. Q(0) means that 0+1 > 2*x, which would mean that 1 > 0. This is true.\\  b. Q(-1) means that -1+1 > 2*-1, which means that 0 > -2. This is true.\\   c. Q(1) means that 1+1 > 2*1, which means that 2 > 2. This is false. \\  d. This says "There exists some integer x such that Q(x) is true". As we have found two previously, namely 0 and -1, this is true. \\  e. This says "For all integers x, Q(x) is true". As we have found an integer that Q(x) is false for, namely 1, this is false. \\  f. This says "There exists some integer x such that Q(x) is not true". As we have found an example, namely x=1, this is true.\\   g. This says "For all integers x, Q(x) is not true". As we have found a counterexample, namely x=0 and x=1, this is false. \\  8)\\  a) There are some students in your class who has sent an email message to some  students in your class.\\  b) There are some students in your class who has sent an email message to every  student in your class.\\  c) Every student in your class has sent an email message to some students in your  class.\\  d) There are some students in your class who has been sent an email message by  every student in your class.\\  e) Every student in your class has been sent an email message by some students in  your class.\\  f) Every student in your class has sent an email message to every student in your class.\\  12)\\  a) Let A(x) be “User x has access to an electronic mailbox.” Then we have $\forall$xA(x).\\  b) Let A(x,y) be “Group member x can access resource y,”and let S(x,y) be “System x is in state y.” Then we have S(file system,locked) $\rightarrow$ $\forall$xA(x,system mailbox)\\  c) Let S(x,y) be “System x is in state y.” Recalling that “only if” indicates a necessary condition, we have S(firewall,diagnostic) $\rightarrow$ S(proxy server, diagnostic). \\  d) Let T(x) be “The throughput is at least x kbps,” where the domain of discourse is positive numbers, let M(x,y) be “Resource x is in mode y,” and let S(x,y) be “Router x is in state y.” Then we have (T(100)$\wedge$$\reg$T(500)$\wedge$$\reg$M((proxy server, diagnostic)) $\rightarrow$ $\exits$xS(x,normal)\\