Are the following statements correctly translated?

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Using predicate symbols shown below and appropriate quantifiers, write each English language statement as a predicate wff.

Domain is all the objects in world.

  1. B(x) : x is a bee
  2. F(x) : x is a flower
  3. L(x,y) : x loves y

Following are the statements along with my attempt at the question. Kindly give a hint if any ( or all :( ) of the following are wrong. An English translation of my attempted solution would be very helpful in case I did it wrong.

  1. All Bees love all flowers

    $\forall$ b $\forall$ f ( B(b) $\wedge$ F(f) $\rightarrow$ L(b,f) )

  2. Some Bees love all flowers

    $\forall$ f $\exists$ b ( B(b) $\wedge$ F(f) $\rightarrow$ L(b,f) )

  3. All Bees love some flowers

    $\forall$ b ( B(b) $\wedge$ ( $\exists$ f ( F(f) $\rightarrow$ L(b,f) )

    or this

    $\forall$ b $\exists$ f ( B(b) $\wedge$ F(f) $\rightarrow$ L(b,f) )

  4. Every bee hates only flowers.

    (This 'only' is particularly causing confusion, should I account for the fact that there are other non-flower objects in domain )

    $\forall$ b $\forall$ f ( B(b) $\wedge$ F(f) $\rightarrow$ $\neg$ L(b,f) )

  5. Every bee hates all flowers

    $\forall$ b $\forall$ f ( B(b) $\wedge$ F(f) $\rightarrow$ $\neg$ L(b,f) )

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There are 2 best solutions below

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  1. All Bees love all flowers

    $\forall$ b $\forall$ f ( B(b) $\wedge$ F(f) $\rightarrow$ L(b,f) )

    Just add parentheses around the antecedent: $\forall b \forall f \Big( (B(b) \wedge F(f)) \rightarrow L(b,f) \Big)$

  2. Some Bees love all flowers

    $\forall$ f $\exists$ b ( B(b) $\wedge$ F(f) $\rightarrow$ L(b,f) )

    Here you need to put $\exists b$ before $\forall f$, and the main connective needs to be $\land$: $$\exists b \Big(B(b) \land \forall f(F(f) \rightarrow L(b, f)\Big)$$

  3. All Bees love some flowers

    $\forall$ b ( B(b) $\wedge$ ( $\exists$ f ( F(f) $\rightarrow$ L(b,f) )

    or this

    $\forall$ b $\exists$ f ( B(b) $\wedge$ F(f) $\rightarrow$ L(b,f) )

    You need $\rightarrow$ as the main connective: $$\forall b \Big(B(b) \rightarrow \exists f(F(x) \land L(b, f))\Big)$$

  4. Every bee hates only flowers.

    (This 'only' is particularly causing confusion, should I account for the fact that there are other non-flower objects in domain ).

    Yes, you need to account for other non-flower objects in the domain. Think of the statement as expressing the equivalent: "All bees hate (i.e. do NOT like) flowers, and for everything ( $\forall x$ ) that is not a flower $(\forall x \lnot F(x))$, then bees love x. I would suggest adding a predicate: $H(x, y):\;$ "x hates y". But if you are given only the predicates you posted, try using $\lnot L(x, y)$ to convey "x does not like y".

    Try taking a "go" at this translation once again.

  5. Every bee hates all flowers

    $\forall$ b $\forall$ f ( B(b) $\wedge$ F(f) $\rightarrow$ $\neg$ L(b,f) )

    This is fine, though you need to put additional parentheses around $B(b) \land F(f)$. And here, too, I'd suggest using a predicate "hates" instead of negating "loves".

3
On

(2) Some Bees love all flowers, i.e. there are some bees $x$ such that (if any $y$ is a flower, $x$ loves $y$), i.e. $\exists x(Bx \land$ (if any $y$ is a flower, $x$ loves $y$)), whence

$\exists x(Bx \land \forall y(Fy \to Lxy))$

[Does your text allow $b$, for example, as a variable? Usually early-alphabet letters in FOL serve as names, a.k.a. constants.]

(3) All Bees love some flowers, i.e. if any $x$ is a bee, then for some flower $y$, $x$ loves $y$, i.e. $\forall x(Bx \to$ (for some flower $y$, $x$ loves $y$)), whence

$\forall x(Bx \to \exists y(Fy \land Lxy))$

I would add that $x$ hates $y$ does not translate as $\neg Lxy$. Not loving something doesn't imply you hate it -- you could be indifferent.