Open main menu

lensowiki β

Changes

Philosophy/PS2

109 bytes removed, 19:29, 13 February 2009
8b
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.
{|
! p || q || r || p .q⊃r || p∨q⊃r || q$pp⊃(q⊃r)|| q⊃(p⊃r) || (p⊃r).(q⊃r) || (p⊃r)∨(q⊃r)
|-
|T || T || T || T || T || T || T || T || T
|-
|T || T || F || F || F || F || F || F || F
|-
|T ||F || T || T || T || T || T || T || T
|-
|T ||F || F || T || F |}(c){|! p T || q T || p$q F || (p$q)$pT
|-
|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
|}
(d)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)∨(p$qr⊃q)$r
|-
| 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. (ea)
{|
! p || q || r || φ(p⊃rp,p,p) || (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 ||
|}
7. (a)
{|
! p || p || r q || φ(p⊃r) || (p⊃p) . (p⊃rq,p,q)
|-
| T || T || T ||
|-
| T || T || F || T
|-
| F || F T || 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 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 || F || |-| T || F || T ||
|-
| T || F T || F T || T
|-
| F T || T || T F || F
|-
| F || T || F T || F
|-
| F || F || T || |-| F || F || F || T
|}
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.
Anonymous user
Retrieved from "http://www.lensovet.net/~sysadmin/w/Special:MobileDiff/3694...3706"