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6.
(a) {|! p || p || p$p|-|T || T || F|-|F || F || F|}(b){|! q || p || q$p|-|T || T || |-|T || F || |-|F || T || |-|F || F || |}(c){|! p || q || p$q || (p$q)$p|-|T || T || |-|T || F || |-|F || T || |-|F || F || |}(d){|! p || q || r || (p$q)$r|-| T || T || T || |-| T || T || F || |-| T || F || T || |-| T || F || F || |-| F || T || T || |-| F || T || F || |-| F || F || T || |-| F || F || F || |}(e){|! p || q || r || (p⊃r) || (p⊃r)$q|-| T || T || T || |-| T || T || F || |-| T || F || T || |-| T || F || F || |-| F || T || T || |-| F || T || F || |-| F || F || T || |-| F || F || F || |} 7. (a){|! p || p || r || (p⊃r) || (p⊃p) . (p⊃r)|-| T || T || T || |-| T || T || F || |-| F || F || T || |-| F || F || F || |}(b){|! p || r || q || (r⊃q) || (p⊃q) || (r⊃q) . (p⊃q)|-| T || T || T || |-| T || T || F || |-| T || F || T || |-| T || F || F || |-| F || T || T || |-| F || T || F || |-| F || F || T || |-| F || F || F || |}(c){|! p || r || q || (p⊃f(p,q,p)) || (q⊃f(p,q,p)) || (p⊃f(p,q,p)) . (q⊃f(p,q,p))|-| T || T || T || |-| T || T || F || |-| T || F || T || |-| T || F || F || |-| F || T || T || |-| F || T || F || |-| F || F || T || |-| F || F || F || |}(d){|! p || r || q || ((p⊃q)⊃r) || ((q⊃p)⊃r) || ((p⊃q)⊃r) . ((q⊃p)⊃r)|-| T || T || T || |-| T || T || F || |-| T || F || T || |-| T || F || F || |-| F || T || T || |-| F || T || F || |-| F || F || T || |-| F || F || F || |} 8.(a) Yes, as the truth-values for s#-t and -s#t are the same.(b){|! s || t || u || (s#t) || (s#t)#u || t#u || s#(t#u)|-| T || T || T || |-| T || T || F || |-| T || F || T || |-| T || F || F || |-| F || T || T || |-| F || T || F || |-| F || F || T || |-| F || F || F || |}
9. (a) Compare the values of the 2nd and 3rd rows; if they are equivalent, then # is commutative. Otherwise, it isn't.
(b) ??Say P and Q are equivalent, i.e. both true. Then, for a connective to be anti-commutative, that means that P#Q is equivalent to -(Q#P). However, this is impossible as P#Q and Q#P are equivalent, so the negation of the latter cannot be equivalent to the former.
10.
(a) No. Say p = "God exists", but there is actually no God. Many people believe God exists is true. Say p = "oceans are blue", which they are. Many people believe oceans are blue is true. Thus the truth value of the statement depends on more than just the truth-value of its single component.
(b)No. Say p = "the sun's gravitational field exerts an attractive force on all objects", q = "some frogs are green"; then the whole statement is false. However, say p remains the same and q = "the planets orbit around the sun"; then, the whole statement is true. Thus the whole statement can be both true and false if p & q are both true, and it is not truth-functional as the truth-values of p & q alone do not determine its truth value.
(c) Yes; this is simply the "not" connective, whose output depends solely on the truth value of the input.