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 (a2) George will cut down a tree to be false, q = T and u = F. (George will marry Martha ∨ George We can try to find an assignment of the other variables that will die a bachelor)make (b1) (Abe will not become mayor true. city will not prosper) ∨ For p ⊃ (Abe will become mayor q⊃r. Ava will become head of chamber of commerce)(cs∨t) (Steve escapes the country ∨ Steve befriends Sally) ⊃ Steve will to be safe(d) true, r = s = t = T. In this case, (-Jerry stays in town p. -Joan reappearsq.∨r.s∨r.t) ⊃ Jan ⊃u becomes false, so the conjunct is false as well. Any other assignments of r, s, and t will triumph make the first part of the conjunct false, thus making the entire conjunct false. Jan will convince JoeThus for all truth-values making (e2) false, (-French object to pact . -Belgians object to pact1) . Italian forces withdraw from Spain . attacks on British ships cease ⊃ Italo-British pact will take effectis false as well, and (f1) implies (-Mail-order campaign breaks Dripsweet monopoly . -mail-order campaign restores competition2) ⊃ Jones will mortgage his home ∨ (Jones will sell his car . Jones will sell his boat)
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
|-
|}
