Question

Sometimes, but not always, quantifiers distribute over logical operations. Determine which of the following pairs of statements are equivalent. In the case of nonequivalent pairs, give an example of propositional functions P(x) and Q(x) for which the paired statements are not equivalent

problems and attempts coming in a bit...

Answer 1

a. (∀x) [P(x) ^ Q(x)] and (∃x)P(x) ^ (∃x)Q(x)
I got. for the left hand side.
All x's with p and q.
For the right hand side
Some x's with p or some x's with q.
Is there a way to tell that they're not equivalent or are equivalent ?
What if I switch P(x) and Q(x) in these problems? Or let P(x) and Q(x) be a function

Answer 2

b. (∃x)[P(x) ^ Q(x)] and (∃x)(P(x)) ^ (∃x)(Q(x))
left: some x's with p and q
right: some x's with p and some x's with q

Answer 3

c.(∀x)[P(x) V Q(x)] and (∀x)P(x) V (∀x)Q(x)
left There exists all values of x with p or q
right There exists all values of P or there exists all values of Q.

Answer 4

d.(∃x)[(P(x) V Q(x)] and (∃x)P(x) V (∃x)Q(x)
There exist some values of x in P or Q
There exist some values of P or there exists some values of Q

Answer 5

( ∀x) [P(x) ---> Q(x) ] and ( ∀x)P(x) ---> ( ∀x)Q(x)
Left: There exists all values of x in P as well as Q.
Right: All x values in P will also have all x values in Q

Answer 6

f. ( ∀x)[P(x)<->Q(x)] and ( ∀x)P(x) <-> ( ∀x)Q(x)
All x values are in p if and only if there's q
All x values are in p if and only if there are all x values in q.

Answer 7

@wio @UnkleRhaukus @inkyvoyd

Answer 8

∀ - for all
∃ - for some
∃! - there is one unique solution

Answer 9

@thomaster

Answer 10

are you asking whether the two are equivalent?

Answer 11

yeah

Answer 12

first you have to write it out to get a better understanding... then you have to determine if they are equivalent...

Answer 13

I don't think the two are equivalent.

Answer 14

ok how can we prove that?

Answer 15

but moreover, is the sentence structure correct to begin with?

Answer 16

I dont think they are equivalent either,

the easiest method to disprove , would be to find a single counter example

Answer 17

i was thinking if you can find an x the P(x) and Q(x) is not true, then the statement on the right side is true, but the one on the left side is false

Answer 18

like we could have P(x) be something. Q(x) be something as well.

Answer 19

this is a tough one....

Answer 20

and then apply to the...........problems... maybe :/

Answer 21

(∀x) [P(x) ^ Q(x)] and (∃x)P(x) ^ (∃x)Q(x)
means For all of the x's P and Q.
but the right one is For some x, there is P and for some x there is q

Answer 22

maybe we could let P(x) and Q(x) be a function... ?

Answer 23

we could. P(x) and Q(x) can represent anything

Answer 24

like let P(x) = 3x+ y = 8 hahah I'm making this up
Q(x) = 4x+ 6 = 10 ???

Answer 25

what about y?

Answer 26

oh sdfjkaljfdl; so make Q(x) = 4x + 6y =10

Answer 27

The problem I was thinking is that does x have to be the same object in the statement (∃x)P(x) ^ (∃x)Q(x)

Answer 28

I know that it's for some x there is P and for some x there is Q

Answer 29

(∃x)P(x) ^ (∃ *anotherP x)Q(x)?

Answer 30

(∃x)P(x) ^ (∃ *another x)Q(x)?

Answer 31

typo :D

Answer 32

like a y?

Answer 33

@agent0smith torture time!

Answer 34

yea, that shows that x and y are not necessarily a same object

Answer 35

hmmmmm I don't get it :/
unless
(∃x)P(x) ^ (∃ y)Q(x)?

Answer 36

so for some x there is a p and for some y there is a q

Answer 37

that doesn't say much unless you make P(x) and Q(x) a simple function or something . or sdfljkad;k

Answer 38

(∃x)P(x) ^ (∃ y)Q(y)

Answer 39

for some value of x there is p and for some value of y there is q

Answer 40

except that we have x and y coming into play

Answer 41

@Isaiah.Feynman

Answer 42

Not here yet. What maths is this?

Answer 43

let x be a member in [-2,2], let p(x) := sqrt(1-x^2) ≥ 0 , and Q(x) = x^2 ≥ 0

then, (∃x)P(x) ^ (∃x)Q(x) is true,
but, (∀x) [P(x) ^ Q(x)] is false

Answer 44

alright so .. for some x there is p and for some x there is q. so we can apply a number into those functions.

Answer 45

yep, that is a good counter example :DD

Answer 46

like let x = 2
p(2) = sqrt ( 1-4) ≥ 0
Q(2) = 4 ≥ 0

Answer 47

:O that's a very big number . I don't see how sqrt(-3) can be greater than 0.

Answer 48

maybe -2 would work better.

Answer 49

P(-2) sqrt(1-(-2)^2) >0
Q(-2) = 4 ≥ 0

Answer 50

Not read the whole question but the only way
p(x) := sqrt(1-x^2) ≥ 0
is true is if x =< 1

Answer 51

oh there is some value... some values of x may work to satisfy p and q. but not all of the values of x would work.

Answer 52

(∀x) [P(x) ^ Q(x)] is all values of x in P and Q but if let P and Q be simple equations and test out all the values of x, something is going to be false. either P(x) or Q(x) maybe both will give out false results.

Answer 53

how can 0 be greater than or equal to 0 ?eh that' s 0 = 0

Answer 54

0 is greater than or EQUAL to zero :P

Answer 55

let x = fractions :D

Answer 56

again, we're trying to disprove (∀x) [P(x) ^ Q(x)] and (∃x)P(x) ^ (∃x)Q(x) are equivalent by find a counter example

Answer 57

which i did

Answer 58

that wasn't me who did that 0 thing.. but you did mention that let x [-2,2]

Answer 59

"let x be a member in [-2,2], let p(x) := sqrt(1-x^2) ≥ 0 , and Q(x) = x^2 ≥ 0 then, (∃x)P(x) ^ (∃x)Q(x) is true, but, (∀x) [P(x) ^ Q(x)] is false"

Answer 60

so I pick any number between -2 and 2 that produces non equivalency or equivalency

Answer 61

p(x) := sqrt(1-x^2) ≥ 0

is true is if -1 =< x =< 1

and Q(x) = x^2 ≥ 0

true if x >= 0

so 0 =< x =< 1

Answer 62

0 is a BAD NUMBER D:

Answer 63

so for some values of x, the conditions of p and q are satisfied... but not values of x can do that.

Answer 64

and I want a pizza now... T_T

Answer 65

well, to show that (∃x)P(x) ^ (∃x)Q(x) is true, let x = 0
sqrt(1-0^2) ≥ 0 is true, and 0^2 ≥ i strue.
however, P(x) and Q(x) are not true for all x in [-2,2]. let x = -2

Answer 66

typo 0^2 ≥0 is true

Answer 67

:S 0^2 is greater than or equal to 0... how D:

Answer 68

0 ≥ 0 is a true statement

Answer 69

remmeber, ≥ means greater OR equal

Answer 70

0 is equal to 0. Therefore it is greater than OR equal to zero.

Answer 71

oh rightttttttttt...

Answer 72

so do I come up with my own P(x) and my own Q(x) for each of these situations?

Answer 73

@satellite73 I know you're here ^^

Answer 74

yeah, you can come up with a different counter example. But the important thing is that do you now see that (∀x) [P(x) ^ Q(x)] and (∃x)P(x) ^ (∃x)Q(x) are not equivalent?

Answer 75

yes I could... if I come up with my own p(x) and q(x) to prove that they're not equivalent by letting x be a number... it's easier to see how these conditions can easily be broken.

Answer 76

so I create my own P(x) function and q(x) function for
(∃x)[P(x) ^ Q(x)] and (∃x)(P(x)) ^ (∃x)(Q(x))

That states that for some x there is p and q and on the right there is some x for p and some x for q

Answer 77

so let P(x) = 5x + 2 = 7 lol maybe >:)
Q(x) = 4x + 1 = 5 XD
x = 1 that might work for
for some x there is p and q

Answer 78

well, you need to state what your domain is

Answer 79

domain?

Answer 80

umm [-1,1] ?

Answer 81

that works

Answer 82

so for this we do have some value of x that works for p and q and some value of x that works for p and some value of x that works for q
while x = -1. 0 doesn't work, x = 1 works :)

Answer 83

yes, 1 works. Everything else in [-1,1] doesn't

Answer 84

c.(∀x)[P(x) V Q(x)] and (∀x)P(x) V (∀x)Q(x)
left There exists all values of x with p or q
right There exists all values of P or there exists all values of Q.
So P(x) and Q(x) must be a function that satisfies all values of x. ANd since this is an or statement, it shouldn't be that hard to test for equivalency

Answer 85

:S I found a typo in my A T___T gawd I feel horrible ... the left side is correct but the right is there exists all values of x for p or there exists all values of x for q. D*****!*!*!(&#(*!&@(*!&(*

Answer 86

(∀x) [P(x) ^ Q(x)] and (∀x)P(x) ^ (∀x)Q(x) changes everything!

Answer 87

d.(∃x)[(P(x) V Q(x)] and (∃x)P(x) V (∃x)Q(x)
There exist some values of x in P or Q
There exist some values of P or there exists some values of Q
Let the domain be [-1.1]. Moreover let P(x) = x+ 2 = 3 and Q(x) = x-3 = -2
If x = 1 then it satisfies P(x) and Q(x).

Answer 88

oops it satisfies P(x) or Q(x) for d. oy

Answer 89

(∀x) [P(x) ^ Q(x)] and (∀x)P(x) ^ (∀x)Q(x) is A. UGH!
That means .. for all values of x in P and Q AND for all values of x in P and for all values of x in Q. so all the x's out there need to satisfy or we can do a contradiction.

Answer 90

hmmm the domain needs to be a bit bigger for d. because this time we can satisfy p OR Q

Answer 91

unless we make P(x) x^2 >0
Q(x) x < 0
from [-1.1] yES! That satisfies one or the other.

Answer 92

d.(∃x)[(P(x) V Q(x)] and (∃x)P(x) V (∃x)Q(x)
There exist some values of x in P or Q
There exist some values of P or there exists some values of Q
Let the domain be [-1.1]. Moreover, let P(x) = x^2 >0 and Q(x) = x <0
If we let x = -1, it doesn't satisfy P, but it satisfies Q
If we let x = 1, it satisfies P, but it doesn't satisfy Q.
SInce we have an or statement, we can choose to satisfy P or Q.

Answer 93

oh wait try... P = x > 0 and Q = x < 0 domain from -1,1

Answer 94

@sourwing

Answer 95

or statements might be easier to do... I just thought of letting P(x) = x^2 > 0 and Q(x) = x^3 > 0
:O I think I figured out my A problem.
for all values of x so let the domain be [-2,2] and plug into these equations

P(x) = x^2 > 0 and Q(x) = x^3 + 9 > 0

it works! because all of the results are greater than 0

Answer 96

(∀x) [P(x) ^ Q(x)] and (∀x)P(x) ^ (∀x)Q(x)
for all values of x there is p and q and for all values of x there is p and for all values of x there is q. THat example alone satisfies this ^

Answer 97

@LastDayWork wassup ^^