6.
(a)p = Smith was the murdererq = Jones was lyingr = Jones met Smith last night{|s = murder took place after midnight! P1: -p || . -q ⊃ rP2: r ∨ -s ⊃ p || P3: -p$. q ⊃ -sC: p|Thus, does (-|T || T || F|p.-q ⊃ r) . (r∨-s ⊃ p) . (-p.q ⊃ -s) imply p?|Say p = F || . To make P3 true, q = T and s = F || F|}. This makes P2 false, so we cannot make the antecedent true. Thus the premises taken together do imply the conclusion.
(b)
{|p = trains stop running! q || p || q$p= airline prices will increase|-|T || T || |-|T || F || r = buses reduce their fares|-t = buses lose customers|F || T || |-|F || F || |}(c){|! P1: p || ⊃ q || p$q || (p$q)$p|P2: -p ⊃ r|T || T || |P3: q ⊃ -t|T || F || C: r ⊃ t|-|F || To make conclusion false, we must set r = T || |-|and t = F || F || |}(d){|! . P2 and P3 will always be true, regardless of what values p || and q || r || take on. Thus it is possible for P1 to be true as well (p$= q)$r|-| T || T || T || |-| T || T || F || |-| T || F || T || |-| T || F || F || |-| F || T || T || |-| F || T || F || |-| F || F || = T || |-| F || F || F || |}(e, for example){|! p || q || r || (p⊃r) || (p⊃r)$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 || |}, and the consequent can be false while the antecedent is true. Thus the premises taken together do not imply the conclusion.
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.