logic

Question

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.

Answer 1

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))

Answer 2

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

Comments

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.

---

k :)

---

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

Answer 3

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) $\exists y \forall x F(x) \wedge P(y)$

Comments

Why y and x in d?

---

For all the friends, there is someone who is perfect.

---

But that's not the question rt? The perfect person must be one among the friends.

---

My mistake, Thank you.

Answer 4

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))