Changes
8b
7.
{|! p || q || r || p.q⊃r || p∨q⊃r || p⊃(aq⊃r)|| q⊃(p⊃r) || (p⊃r).(q⊃r) || (p⊃r)∨(q⊃r)|-|T || T || T || T || T || T || T || T || T|-|T || T || F || F || F || F || F || F || F|-|T || F || T || T || T || T || T || T || T|-|T || F || F || T || F || T || T || F || T|-|F || T || T || T || T || T || T || T || T|-|F || T || F || T || F || T || T || F || T|-|F || F || T || T || T || T || T || T || T|-|F || F || F || T || T || T || T || T || T|}Thus we can see that #1, 2, and 4 are equivalent to p.q⊃r while #3 is equivalent to p∨q⊃r
{|! p || q || r || r⊃p.q || r⊃p∨q || (r⊃p).(r⊃q) || (r⊃p)∨(r⊃q)|-|T || T || T || T || T || T || T|-|T || T || F || T || T || T || T|-|T || F || T || F || T || F || T|-|T || F || F || T || T || T || T|-|F || T || T || F || T || F || T|-|F || T || F || T || T || T || T|-|F || F || T || F || F || F || F|-|F || F || F || T || T || T || T|}Thus we can see that #1 is equivalent to r⊃p.q and #2 is equivalent to r⊃p∨q 8. (a){|! p || φ(p,p,p)|-|T || T|-|F || F|} {|! p || q || φ(q,p,q)|-|T || T || T|-|T || F || T|-|F || T || T|-|F || F || F|} {|! p || φ(q,p,q) || q || φ(p,φ(q,p,q),q)|-|T || T || T || T|-|T || T || F || F|-|F || T || T || F|-|F || F || F || T|} (b) Yes; for φ(p,p,p) = F, p = F as well.Yes; for φ(q,p,q) = F, p = F as well.No; there are two cases in which φ(p,φ(q,p,q),q) = F, and p = F for only one of them. Thus p = T, q = F gives us φ(p,φ(q,p,q),q) = F while p = T, so implication does not hold.