1. The schemata schema is not valid if we can find just one assignment of truth-values for which it is false.(a) p = T, q = F, r = T makes the schemata schema false. Not valid.(b) The only way to make this false is for p ⊃ r to be false and p ⊃ q.r to be true. This fixes p to T and r to F. However fixing r to F makes p ⊃ q.r false as well (regardless of the value of q), so it is impossible to make the antecedent true and consequent false here. Thus schemata schema is valid.(c) Making the schemata schema false requires each all of the disjuncts being to be false. To make first one false, fix p to T and q to F (or the opposite, makes no difference). To make second one false, r must be made F since p is already fixed. Thus r and q have the same truth-value and the last disjunct evaluates to true. Again, impossible to make an assignment of truth-values giving us a false statement, so schemata schema is valid.

3. Neither the President (a) No. p ⊃ q is false if p = T and q = F, but (p⊃r)⊃(q⊃r) is happy nor true with those assignments.(b) No. p ⊃ q is Congress placated; The President false if p = T and q = F, but (r⊃p)⊃(r⊃q) is not happy or Congress true with those assignments (for example with r = F).(c) Yes. q = T, r = F, p = T makes the second schema false, and is placatedthe 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|}

(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.

! 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

(e)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).(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

78. (a)

! 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 ||

| T || T || F || T

| T || F || T || 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 ||

! 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

| T || F || F || |-| F || T || T || |-| F || T || F || |-| F || F || T || |-| F || F || F ||

9. (ab) Compare the values of the 2nd and 3rd rowsYes; if they are equivalentfor φ(p,p, then # is commutative. Otherwisep) = F, it isn'tp = F as well.Yes; for φ(b) Say P and Q are equivalentq, i.e. both true. Thenp, for a connective to be anti-commutative, that means that P#Q is equivalent to -(Q#Pq). However= F, this is impossible p = F as P#Q and Q#P well.No; there are equivalenttwo cases in which φ(p, so the negation of the latter cannot be equivalent to the former. 10.φ(aq,p,q),q) No. Say p = "God exists"F, but there is actually no God. Many people believe God exists is true. Say and p = "oceans are blue", which they are. Many people believe oceans are blue is trueF for only one of them. 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"T, q = "some frogs are green"; then the whole statement is false. HoweverF gives us φ(p, say p remains the same and φ(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 "= F while p = T, so implication does not" connective, whose output depends solely on the truth value of the inputhold.