Friday, April 16, 2010

how to prove it - ch1, sec1.5(the conditional and biconditional connectives) ex

This section introduces the Conditional() and the Biconditional() logical connectives and a form of valid reasoning, which is

PQ,PQ

Another important thing to learn is, what these connectives usually translate in English to, Conditional can translate to

If P then Q
Q, if P
Q, when P
Q unless ¬P
P only if Q
P is a sufficient condition for Q
Q is a necessary condition for P

and Biconditional can translate to

P if and only if Q
P iff Q
P is the necessary and sufficient condition for Q

Some important equivalences...

PQ = ¬PQ = ¬Q¬P (this last one is known as contrapositive law)

PQ = PQQP

=========================================================================================

Ex-1(a)
S¬E¬H where
S = Unpleasant smell
E = Explosive
H = Hydrogen

Ex-1(b) FHD

Ex-1(c) FHD
= ¬(FH)D
= (¬F¬H)D
= (¬FD)(¬HD)
= (FD)(HD)

Ex-1(d) A(OP) or (AO)P
A means x2
O means x is odd
P means x is prime


Ex-2(a) SPA
S means, Marry will sell her house
P means, she can get a good price
A means, she finds a nice apartment

Ex-2(b) MCD
C means, having a good credit history
D means, an adequate down payment
M means, getting the mortgate

Ex-2(c) Given sentence is equivalent to, If someone does not stop John then he'll kill himself. Its logical form is.. ¬SK

where S means, someone stops john and
K means, john kills himself

Ex-2(d) D(x,4)D(x,6)¬P(x)
where D(x,y) means, x is divisible by y and
P(x) means x is prime

Ex-3
(a) RW¬S
R means, its raining
W means, its windy
S means, sun is shining

(b) W¬SR , clearly its the converse of (a)

(c) RW¬S, its same as (a)

(d) W¬SR , clearly its the converse of (a)

(e) S¬W¬R
= ¬(¬SW)¬R
= R¬SW
= RW¬S , same as (a)

(f) (RW)(R¬S)
= (¬RW)(¬R¬S)
= ((¬RW)¬R)((¬RW)¬S))
= ¬R((¬RW)¬S))
= ¬R((¬R¬S)(W¬S))
= (¬R(¬R¬S))(W¬S)
= ¬R(W¬S)
= RW¬S, same as (a)

(g) (WR)(¬SR)
= (¬WR)(SR)
= (¬WS)(RR)
= ¬(W¬S)R
= W¬SR, converse of (a)

Ex-4(a)
S means, Sales will go up
E means, Expenses will go up
H means, Boss will be Happy

Logical form of given statement is (SE)(SH)(E¬H) and ¬(SE)















SEHSESHE¬H(SE)(SH)(E¬H)¬(SE)
FFFFTTFT
TFFTFTFT
FTFTTTTT
TTFTFTFF
FFTFTTFT
TFTTTTTT
FTTTTFFT
TTTTTFFF


Clearly all the places where (SE)(SH)(E¬H) is T, ¬(SE) is also T . So, ¬(SE) can be inferred from (SE)(SH)(E¬H) and the argument is valid.

Ex-4(b)
X means, Tax rate goes up
U means, Unemployment rate goes up
R means, there will be a recession
G means, GNP goes up

logical form of the given statement is (XUR)(G¬R)(GX), ¬U























XURGXURG¬RGX(XUR)(G¬R)(GX)¬U
FFFFTTFFT
TFFFTTFFT
FTFFTTFFF
TTFFFTFFF
FFTFTTFFT
TFTFTTFFT
FTTFTTFFF
TTTFTTFFF
FFFTTTFFT
TFFTTTTTT
FTFTTTFFF
TTFTFTTFF
FFTTTFFFT
TFTTTFTFT
FTTTTFFFF
TTTTTFTFF


using the same argument as earlier, its a valid argument.

Ex-4(c)
W means, Warning light will come on
P means, Pressure is too high
R means, Relief valve is clogged

logical form is (WPR)¬R , WP















WPRPRWPRWP
FFFFTT
TFFFFF
FTFFTF
TTFFFT
FFTFTT
TFTFFF
FTTTFF
TTTTTT


we can see an inconsistency at line 3, so its not a valid argument.


Ex-5(a) PQ
= (PQ)(QP)
= (¬PQ)(¬QP)
= ((¬PQ)¬Q)((¬PQ)P)
= ((¬P¬Q)(Q¬Q))((¬PP)(QP))
= (¬P¬Q)(QP)
= (PQ)(¬P¬Q)

Ex-5(b) (PQ)(PR)
= (¬PQ)(¬PR)
= (¬P¬P)(QR)
= ¬P(QR)
= P(QR)


Ex-6(a) (PR)(QR)
= (¬PR)(¬QR)
= ((¬PR)¬Q)((¬PR)R)
= ((¬PR)¬Q)R
= ((¬P¬Q)(R¬Q))R
= (¬(PQ))((R¬Q)R)
= (¬(PQ))R
= PQR

Ex-6(b) (PR)(QR) = PQR
Here is the proof...
(PR)(QR)
= (¬PR)(¬QR)
= (¬P¬Q)(RR)
= ¬(PQ)R
= PQR


Ex-7(a)














PQRPQQPQRRQ(PQ)(QR)PQRQ(PQ)(RQ)PR(PR)((PQ)(RQ))
FFFTTTTTTTTTT
TFFFTTTFFTTFF
FTFTFFTFFFFTF
TTFTTFTFTFTFF
FFTTTTFTTFTTT
TFTFTTFFFFFTF
FTTTFTTTFTTTT
TTTTTTTTTTTTT


Match column 8 and 13, clearly both are same.

Ex-7(b)
(PQ)(QR)
= (¬PQ)(¬QR)
= ¬P(Q¬Q)R
= ¬P(TRUE)R
= TRUE

Ex-8 PQ
= ¬(¬P¬Q)
= ¬(P¬Q)

Ex-9 PQ
= (PQ)(QP) , now using Ex-8
= ¬((PQ)¬(QP))


Ex-10

(a) P(QR)
= ¬P(¬QR)
= ¬P¬QR

(b) Q(PR)
= ¬Q(¬PR)
= ¬Q¬PR
= ¬P¬QR , clearly its same as (a)

(c) (PQ)(PR)
= (¬PQ)(¬PR)
= ((¬PQ)¬P)((¬PQ)R)
= (¬P)((¬PR)(QR))
= ¬P(¬PR)(QR)
= (¬P(¬PR))(QR)
= ¬P(QR)

we've reduced original form to disjunctive normal form(disjunction of conjunctions) here

(d) (PQ)R
= ¬(PQ)R
= ¬P¬QR
clearly its same as (a) and (b)

(e) P(QR)
= ¬P(QR)
clearly its same as (c)

4 comments:

  1. for question 4A how did you come up with that logical statement?

    ReplyDelete
  2. 4(a)

    S means, Sales will go up

    E means, Expenses will go up

    H means, Boss will be Happy

    From the question, take each sentence one by one..

    Either sales or expenses will go up: (SE)
    If sales go up then boss will be happy: (SH)
    If expenses go up then boss will be unhappy: (E¬H)

    Since question means to say all of the above, so we take the conjunction of all of the above logical statements to get (SE)(SH)(E¬H)

    therefore sales and expenses will not both go up, meaning(yes, there is scope of ambiguity here I guess) Both of them together will not go up..so.. not of (both sales and expense going up together): ¬(SE)

    Now, to be honest, I can't really remember how I came up with it but maybe I did reverse engineer a little bit.

    ReplyDelete
  3. Here's an alternative for 7 a)

    (P→Q)∧(Q→R)=
    (P→Q)∧(Q→R)∧[(R∨¬R)∨(¬P∨P)]
    (P→Q)∧(Q→R)∧[(¬P∨R)∨(P∨¬R)]
    (P→Q)∧(Q→R)∧[(¬P∨R)∨(P∨(¬Q∧Q)∨¬R)]
    (P→Q)∧(Q→R)∧[(¬P∨R)∨(¬Q∨P)∧(¬R∨Q)]
    (P→Q)∧(Q→R)∧(P→R)∨(Q→P)∧(R→Q)
    (P→R)∧(P→Q)∧(Q→P)∨(Q→R)∧(R→Q)
    (P→R)∧[P↔Q)∨(R↔Q)]



    ReplyDelete
  4. Hi Himanshu! Thanks for these incredibly helpful posts.

    On Exercise 1.d. shouldn't the correct logical form be A ∧ P → O. Disregarding A for instance, O is by no means sufficient condition for P.

    ReplyDelete