Thanks @Manis :)

Dark Mode

8,165 views

39 votes

Which one of these first-order logic formulae is valid?

- $\forall x\left(P\left(x\right) \implies Q\left(x\right)\right) \implies \left(∀xP\left(x\right)\implies \forall xQ\left(x\right)\right)$
- $\exists x\left(P\left(x\right) \vee Q\left(x\right)\right)\implies \left(\exists xP\left(x\right)\implies \exists xQ\left(x\right)\right)$
- $\exists x\left(P\left(x\right) \wedge Q\left(x\right)\right) \iff \left(\exists xP\left(x\right) \wedge \exists xQ\left(x\right)\right)$
- $\forall x \exists y P\left(x, y\right)\implies \exists y \forall x P\left(x, y\right)$

0

57 votes

Best answer

(**A**) is the answer

- LHS: For every x, if P holds then Q holds

RHS: If P(x) holds for all x, then Q(x) holds for all x.

LHS implies RHS but RHS does not imply LHS.

- LHS: An x exist for which either P(x) is true or Q(x) is true.

RHS: If an x exist for which P(x) is true then another x exist for which Q(x) is true.

LHS does not imply RHS, but on RHS if we change ∃xP(x) to ~∃xP(x), implication becomes TRUE.

- LHS: There exist an x for which both P(x) and Q(x) are true.

RHS: There exist an x for which P(x) is true and there exist an x for which Q(x) is true.

LHS implies RHS but RHS does not imply LHS as the 'x' for P and Q can be different on the RHS

- LHS: For every x, there exist a y such that P(x, y) holds.

RHS: There exist a y such that for all x P(x, y) holds.

Here RHS implies LHS but LHS does not imply RHS as the y on LHS can be different for each x.

2

0

11 votes

(A) LHS= if anyone is robbed then he is searched

RHS= if anyone are robbed then anyone is searched'

Both are not true.

(B) LHS=there is some wrestler or batminton player

RHS= if there is some wrestler then there also have some batminton player

Both are not similar

(C)LHS=there is some player who is wrestler and batminton player

RHS=there is a player who is wrestler and there also has a batminton player

it is true in unidirection but not both direction. like

∃x(P(x) ∧ Q(x)) => (∃xP(x) ∧ ∃xQ(x))

(D)LHS= For all wrestler there is a trainer

RHS= There is same trainer for all wrestler

not true always

edited
Jan 3, 2018
by Puja Mishra

Lets give an example of a class of boy (for ease of writing) students.

$P(x) : x \text{ performs well}$

$Q(x): x \text{ gets medal}$

$\forall x, P(x) \implies Q(x)$

means,

- For everyone belonging to class, if he performs well then he gets medal
- Equivalently, if anyone in the class performs well, then he gets a medal

$(\forall x P(x)) \implies \forall x Q(x)$

means,

- If everyone in the class performs well then each of them gets a prize.
- This is different from the previous as now the medal requires a good performance from everyone in the class and not individual performance.
- Still the first statement do imply the second one because in first case we say that anyone performing well is given a medal. So, even if everyone performs well, every one will get medal.

11

option d can be termed as x+y=0. Domain is the real number

LHS: for all x there exist some y which makes x+y=0 is true(1) always as

( x=2,y=-2,x+y=0

x=3,y=-3,x+y=0 and so on.)

RHS: for some y there exist all x which makes x+y=0 is not true(0).

as for (y=2, there is only one value of x=-2 which makes X+Y=0 but other than x=2 , everything else will not make X+Y=0.)

So (LHS=1)->(RHS=0) makes it invalid

LHS: for all x there exist some y which makes x+y=0 is true(1) always as

( x=2,y=-2,x+y=0

x=3,y=-3,x+y=0 and so on.)

RHS: for some y there exist all x which makes x+y=0 is not true(0).

as for (y=2, there is only one value of x=-2 which makes X+Y=0 but other than x=2 , everything else will not make X+Y=0.)

So (LHS=1)->(RHS=0) makes it invalid

0

10 votes

For the options, I treat x for boy, y for girl.

P(x) means, he flirts. Q(x) means he loves.

P(x,y) means the boy & girls both are in relationship. (Prefer this trick or build your own example)

Option A: For all boys ( a boy flirts -> he loves too) -> (all boys flirts -> all boys love) It is **valid**.

Option B: For Some boys ( either flirts or loves) -> ( if a guy flirts he loves too) No way, its **not valid**.

Option C: For Some boys ( both flirts and loves) < -> ( a guy is there who flirts & a guy is there who loves) I think it should be -> not <-> Hence **not valid**.

Option D: For all boys there are some girlfriends -> there are some girls for which all the boys are boy friends. (**Not valid**.)