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+1 vote
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Translate each of these statements into logical expressions
using predicates, quantifiers, and logical connectives.
a) No one is perfect.
b) Not everyone is perfect.
c) All your friends are perfect.
d) At least one of your friends is perfect.
in Mathematical Logic by Boss (20k points) | 1.1k views

4 Answers

+2 votes

a) No one is perfect. == Not ( one is perfect) = ~ (∃x(px))= ∀x ~p(x)= Every one is imperfect.
b) Not everyone is perfect.== Not (everyone is perfect.)= ~( ∀x(px))=∃x ~p(x)= Atleast one is imperfect.
c) All your friends are perfect. == if there is a person who is your friend then he is perfect== ∀x( F(x)→P(x))
d) At least one of your friends is perfect. == There is a person who is your friend who is perfect.

∃x (F(x)∧P(x))

by Veteran (62.7k points)
0 votes

plz correct me if wrong!!

P(x) : perfect

F(x) :friends

(a)∽∃x(P(x))

(b)∽∀x(P(x))

(c)∀x(F(x)------>P(x))

(d)  i am thinking in this way 

ALL FRIENDS ARE PERFECT - NO FRIEND IS PERFECT

 how to write above sentence...?? in logic

by Boss (20k points)
+1

part d) can be thought in this way "there exists a friend of yours that is perfect". In logic, it will be: ∃x(F(x) ^ P(x))

0
it tells that there is only one friend but there may be 2 3 4 5 6 friends and so on which are perfect so how to include that in above logic...??
0

nopes. ∃x(P(x)) can also be read as:

1)there is atleast one x such that P(x) is true 

2) for some x, P(x) is true.

0
k :)
0

is there any answer for this to write in logic"ALL FRIENDS ARE PERFECT - NO FRIEND IS PERFECT"

0 votes
Equivalent answers,

$P(x):$ X is Perfect
$F(x):$ X is a friend of mine

(a) $\forall x \neg P(x)$
(b) $\exists x \neg P(x)$
(c) $\forall x,  F(x) \rightarrow P(x)$
(d) $\exists y \forall x F(x) \wedge P(y)$
by Junior (985 points)
0
Why y and x in d?
0
For all the friends, there is someone who is perfect.
0
But that's not the question rt? The perfect person must be one among the friends.
0
My mistake, Thank you.
0 votes
P(x) : perfect

F(x) :friend

a) ∽∃x(P(x))

B)  ∽∀x(P(x))

C) ∀x(F(x)->P(x))

D) ∃x(Friend(x) ^ Perfect(x))
by Boss (41.5k points)

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