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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$M((proxy (T(100)$\wedge$$\neg$T(500)$\wedge$$\neg$M((proxy  server, diagnostic)) $\rightarrow$ $\exits$xS(x,normal)\\ $\exists$xS(x,normal)\\  16)\\  Let a and b be odd integers. By definition of odd we have that a = 2n + 1 and b = 2m + 1. Consider the sum a + b = (2n + 1) + (2m +1) = 2n + 2m +2 = 2k, where k = n + m + 1 is an integer. Therefore by definition of even we have shown that a + b is even.\\  20)\\  Let r and w be two rational numbers. By definition of rational r= $\frac{a}{b}$ and w=$\frac{c}{d}$ where a,b,c,d are integers and b$\ne$0, d$\ne$0.  Consider r+w= $\frac{a}{b}$ + $\frac{c}{d}$ = $\frac{ad+bc}{bd}$ where bd$\ne$0 is an integer and ad +bc is an  integer. Therefore the sum is rational by definition.\\  24)\\