Changes
8b
(c) Yes. q = T, r = F, p = T makes the second schema false, and is the only assignment that does. It also makes the first schema false, so the first implies the second.
4.For (2) to be false, q = T and u = F. We can try to find an assignment of the other variables that will make (1) true. For p ⊃ (q⊃r.(s∨t)) to be true, r = s = t = T. In this case, (-p.q.∨r.s∨r.t)⊃u becomes false, so the conjunct is false as well. Any other assignments of r, s, and t will make the first part of the conjunct false, thus making the entire conjunct false. Thus for all truth-values making (2) false, (1) is false as well, and (1) implies (2).
5.First schematize the statements.p = prices are lowq = sales are highr = you sell quality merchs = your customers are satisfied (ai){|! becomes (p || ⊃ q || r || p) .-(r || p.q || p.q ⊃ r || s); (ii) becomes (p.-∨ r ∨ ) ⊃ (p.q ⊃ r∨ s).|-|T || (ii) is false when p∨r = T || T || and q∨s = F || T || T || T|-|T || T || F || T || T || F || T|-|T || F || T || F || F || T || T|-|T || F || F || T || F || T || T|-|F || T || T || F || F || T || T|-|F || T || F || F || F || T || T|-|F || F || T || F || F || T || T|-|F || F || F || F || F || T || T|}(b){|! p || q || r || s || -p.-q || This means that q.-= s || = F and p.-!= r.s || -p.-q ∨ q.-s ∨ p.-r.and s|-| T || T || T || T || F || F || F || F|-| T || T || T || F || F || T || F || T|-| T || T || F || T || F || F || T || T|-| T || T || F || F || F || T || F || T|-| T || F || T || T || F || F || F || F|-| T || F || T || F || F || F || F || F|-| T || F || F || T || F || F || T || T|-| T || F || F || F || F || F || F || F|-| F || T || T || T || F || F || F || F|-| F || T || T || F || F || T || F || T|-| F || T || F || T || F || F || F || F|-| F || T || F || F || F || T || F || T|-| F || F || T || T || T || F || F || T|-| F || F || T || F || T || F || F || T|-| F || F || F || T || T || F || F || T|-| F || F || F || F || T || F || F || T|}being false makes (ci){|! false regardless of the values of p || q || r || (p∨q)≡-and r || p∨, so (-q ⊃ ri) || implies (p∨qii)≡-r . p∨(-q ⊃ r)|-|T || T || T || F || T || F|-|T || T || F || T || T || T|-|T || F || T || F || T || F|-|T || F || F || T || T || T|-|F || T || T || F || T || F|-|F || T || F || T || T || T|-|F || F || T || T || T || T|-|F || F || F || F || F || F|}
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$. q ⊃ -sC: p|Thus, does (-|T || T || F|p.-q ⊃ r) . (r∨-s ⊃ p) . (-p.q ⊃ -s) imply p?|Say p = F || . To make P3 true, q = T and s = F || F|}. This makes P2 false, so we cannot make the antecedent true. Thus the premises taken together do imply the conclusion.
(b)
p = trains stop running
q = airline prices will increase
r = buses reduce their fares
t = buses lose customers
P1: p ⊃ q
P2: -p ⊃ r
P3: q ⊃ -t
C: r ⊃ t
To make conclusion false, we must set r = T and t = F. P2 and P3 will always be true, regardless of what values p and q take on. Thus it is possible for P1 to be true as well (p = q = T, for example), and the consequent can be false while the antecedent is true. Thus the premises taken together do not imply the conclusion.
7.
{|
! q || p || q$p|-|T r || T p.q⊃r || p∨q⊃r |-|T p⊃(q⊃r)|| F q⊃(p⊃r) || |-|F || T || |-|F || F || |}(cp⊃r){|! p || q || p$q || .(p$qq⊃r)$p|-|T || T || |-|T || F || |-|F || T || |-|F || F || |}(dp⊃r){|! p || q || r || ∨(p$qq⊃r)$r
|-
| T || T || T || T || T || T || T || T || T
|-
| T || T || F || F || F || F || F || F || F
|-
| T || F || T || T || T || T || T || T || T
|-
| T || F || F || T || F || T || T || F || T
|-
| F || T || T || T || T || T || T || T || T
|-
| F || T || F || T || F || T || T || F || T
|-
| F || F || T || T || T || T || T || T || T
|-
| F || F || F || T || T || T || T || T || T
|}
{|
! p || q || r || r⊃p.q || r⊃p∨q || (r⊃p).(p⊃rr⊃q) || (p⊃rr⊃p)∨(r⊃q)$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
{|
! p || φ(p,p,p || r || (p⊃r) || (p⊃p) . (p⊃r)|-| T || T || T ||
|-
| T || T || F ||
|-
| F || F || T || |-| F || F || F ||
|}
{|
! p || r || q || φ(r⊃q) || (p⊃q) || (r⊃q) . (p⊃qq,p,q)
|-
| T || T || T ||
|-
|-
|-
|}
{|
! s p || t φ(q,p,q) || u q || φ(s#t) || p,φ(s#tq,p,q)#u || t#u || s#(t#u,q)
|-
| T || T || T || T
|-
| T || T || F || F
|-
| T F || F T || T || F
|-
|}
