MAT1348 DISCRETE MATHEMATICS FOR COMPUTING DGD 3
Q1. Use an appropriate truth tree to answer each of the following questions:
i. Is the set�a ^ (b ! c), a $ ¬(d ^ e), ¬
⇥(a _ ¬d) _ (b $ c)
⇤ consistent?
If so, give all truth assignments that justify your claim.
ii. Is X : (p ! r) _ (q $ ¬s) logically equivalent to Y : (p ^ ¬r) ! ¬(q _ s) ?If not, give all counterexamples (that is, give all truth assignments that certify that Xis not logically equivalent to Y ).
iii. Is ¬�¬(a _ b) ! (x _ y)
�! ¬(¬x ^ y) a tautology, contradiction, or contingency?
Q2. Strolling along on The Island of Knights and Knaves, we meet an inhabitant called A(whom is a knight or a knave).
A says: “There is gold on The Island of Knights and Knaves if and only if I am a knight.”
Is there gold on The Island of Knights and Knaves?
hint: break the problem down into two cases based on A’s type. (proof by cases)
Q3. Prove the following theorem in two ways:
i. By Contradiction.
ii. With an Indirect Proof (by Contraposition).
Theorem. Let n be an integer. If n3 + 5 is odd, then n+ 2 is even.
Q4. Consider the following propositions:P : You pass 1348.S : You study for 1348.L : You learn.C : You get confused.H : You get 100% on the final exam.A : You get at least 40% on the final exam.
4.i). Using the above propositional variables, translate the following argument into proposi-tional logic:
You study for 1348, but you get confused.You learn whenever you study for 1348.Getting at least 40% on the final is a necessary condition for you to pass 1348.Getting 100% on the final is a sufficient condition for you to pass 1348.Therefore,
⇥You pass 1348 and get confused
⇤if and only if
⇥you study 1348 and get at
least 40% on the final.⇤
4.ii). Using a truth tree, determine whether the above argument is valid. If it is not valid, giveall counterexamples.
1
Translation
as→ L
P → A
H→p
FEAT
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an ( b → c)
taintedas→c✓m7 A
7 (dnejr%1and )7 ( bbc )
HC
mdTe 7d ie
*:&"¥ "¥ "¥
since all paths are dead ( closed ),
the root can never be True.
°o° there is no truth assignment that makes all three propositions True
% the set is inconsistent .
ii. Is X : (p ! r) _ (q $ ¬s) logically equivalent to Y : (p ^ ¬r) ! ¬(q _ s) ?If not, give all counterexamples
5
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'
7(q←ns)✓
Pmrvplnrv/ \
77(qvs)✓
7
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when p=T ( which means # ' list )
q=T7¥←Y) is Truer=Fs=T When p=T
(which means X←YisF) q=Fr=F ConclusionS=T
@hichmeansxcsyisfj¥11 ogicallyequivalenttoy .
Counterexamples
:X←oYisFlhenanotatautology )
wnenya;E¥yrCjt¥¥)orC¥I¥t
iii. Is ¬�¬(a _ b) ! (x _ y)
�! ¬(¬x ^ y) a tautology, contradiction, or contingency?
6
c_ipropohnorshort.ro#7fi(avbtKvyDt7GxnY@77f(avbtKvyDrTGXAY)1 :
7(avb)→Kvy)✓ giventhat there is already at
/ \ least one complete open path , we
77(avb)vXVY
' Know the proposition at the root ( p )
1 / \ Can be True . ( which means we don't
avbvxvyr needtocompletetherestofthetreej/ \
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, the root 7 Pcanneverbet.
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Q5. Strolling along on The Island of Knights and Knaves, we meet an inhabitant called A(whom is a knight or a knave).
A says: “There is gold on The Island of Knights and Knaves if and only if I am a knight.”
Is there gold on The Island of Knights and Knaves?
hint: break the problem down into two cases based on A’s type. (proof by cases)
7
.
Either A is a Knight or A is a Knave
Cad .Assume A is a Knight .
Then A 's statement must be T since Knights never lie.
A says : (There is gold) a (A is a Knight). -
? T in this case
↳ to make A 's statement T, (There is gold ) must be T
. ( because (T←T) is T ) .
80 IF A is a Knight,then there is gold on The Island .
Cast . Assume A is a Knave.
Then ( in this case ) ,A 's statement must be F since Knaves always lie
A says : (There is gold) a (A is a Knight). -
? F in this case
↳ to make A 's statement F, (There is gold ) must be T
. ( because ( To F) is F)80 IF A is a
Knave,
then there is gold on The Island .
In both possible cases,
it follows that there is gold oh The Island OF K&K .
so there is gold on the Island .
MAT1348C DISCRETE MATHEMATICS FOR COMPUTING T.A. ANDREW WAGNER
DGD 4 ( February 14 )
Q1. Prove the following theorem in two ways:
Theorem. Let n be an integer. If n3 + 5 is odd, then n+ 2 is even.
1i. By Contradiction.
1
Effeteness
Madam speeders
Hk
T #To prove P→Q
To prove P→Q by contradiction ,we assume 7(P→Q) is True
.
Since 7(p→Q)=PmQ ,we start by assuming PMQIST .
Assume P :" n3t5 is odd
"
is true and assume 7Q : "ht2isodd " is True . (goalistoshow contradiction)
since n3t5isodd,
there exists an
|
Since nt2 is odd', there exists an integer
integer k such thatl such that
n3t5=2h+|.
n+2=2l+1
Consequently , 5=2k+1-n3 and n=2lH-2=2tl
Thus
5t2kHf2l-D3-2kHt@eP-3Hej2t3HltD-2kHf8l3-l2l2t6l-D-2kH-8l3t12l2-6lH-2k-8l3H2l2-6lt2-2fr-4l3t6l2-3ltD-2mwherem-k-4l3t6l2-3ltt.ThusmErZ.BydeF.thismeansthat5isevenBThisisacontradictionsinaweKnow5-2tHHisodd.Thereforeassuming7PsQ.isTrue1eadstocontradiction8o7CPsQ1mustberalseInotherwords.P
→ Q( thetheoremjistrue
-
Theorem. Let n be an integer. If n3 + 5 is odd, then n+ 2 is even.
1ii. With an Indirect Proof (by Contraposition).
2
npn-
To prove P→Q.
Q
To prove P→Q with an indirect proof ,we prove 7Q→7P( the contrapositive 7Q→7Pis=P→Q )
7Q : nt2 is odd. 7p:n3+5 is even
.
Assume ) Q is True .ie Assume ht2 is odd. (goal is to prove 7p : n3t5 is even)
Then nt2=2mH for some integer m .
⇒ n=2mtl -2
=2m -1
Consequently , n3+5=(2m - 1)3+5
=(GmP - 3(2m)2t3(2m) - 1) +5
=8m3 - 12m2+6mt4
=2[4m3-6m2t3m+2]
=2j for j=4m3-6m2t3mt2 . thus je 'Z
Thus, bydef.
n3t5 is even . ie 7P is True .
Therefore,
we proved 7Q→7P is true .
=p→÷
2.ii). Using a truth tree, determine whether the above argument is valid. If it is not valid, giveall counterexamples.
2
Q4Donpage1 .
Hoppa
✓
✓root
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1 openpalh ,it means the rootSv( which is the negation OF thecvargument) can be True
.
/ \ Thus,the argument itself can bet
.
75 L ✓ which means the argument is
.%#./\ invalidIFA
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at
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ax
rootcnegationof arguments root ¥0. ;#i
is True when (negation of arguments
S=T is True when
A=TS=T
p=FA=T
H= F
p=FL=T
H=FC= T
L=T* Both the open paths happen to
C=Tgive the same counterexample