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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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$2$ is incorrect, it should be $\exists b:\forall f(B(b)\land F(f)\to L(b,f))$ –  abiessu Feb 8 at 14:26

2 Answers 2

up vote 4 down vote accepted
  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".

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Could you please tell me what my solution for " Some Bees love all flowers" actually meant ? –  Mohammad Yaseen Feb 8 at 14:35
1  
What your translation actually means that for every bee, there is some flower that it likes (for one bee, it might be roses. For another bee, it might be pink carnations). The point being: the liked flowers can vary from bee to bee. What you need to translate, is that there exist flowers that ALL bees like, across the board, so that perhaps those existent flowers are lilacs so that ALL bees like lilacs. Here, the liked flowers do not vary from bee to bee. –  amWhy Feb 8 at 14:44
    
So, I got three wrong, That's sad. Though the suggestion is very good, I am not allowed to add my own predicate. –  Mohammad Yaseen Feb 8 at 14:54
    
That's fine then. Just use $\lnot L(x, y)$ to denote "x hates y", as you intended. –  amWhy Feb 8 at 14:59

(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.

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yes, there is no restriction on variable names. We're using this textbook in class. –  Mohammad Yaseen Feb 8 at 14:40
    
Your explanation is very helpful, thanks for that. –  Mohammad Yaseen Feb 8 at 14:49
    
"Not loving something doesn't imply you hate it" This is indeed correct, But how else can I convey the sense of hating, considering that I can't add my own predicate (say H(x,y) for hate) –  Mohammad Yaseen Feb 8 at 14:51

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