Every bee hates all flowers

Question

I'm trying to translate the following statement to predicate logic:

Every bee hates all flowers.

Let

• $B(x)$: $x$ is a bee.

• $F(x)$: $x$ is a flower.

• $L(x,y)$: $x$ loves $y$.

I came up with these two solutions:

$$(\forall x)[B(x) \to (\forall y)(F(y) \to L’(x,y))]$$

and

$$(\forall x)[F(x)→(\forall y)(B(y) \to L’(y,x))]$$

I feel as though they both say the same thing, but I'm not so sure. The first says

for all things $x$, if $x$ is a bee, and for all things $y$, $y$ is a flower, then $x$ hates $y$.

The second says

for all things $x$, if $x$ is a flower, and for all things $y$, $y$ is a bee, then $y$ hates $x$.

Are they both correct?

Answer 1

Great intuition, sandwich610! Yes, they are correct and logically equivalent. Proof of their equivalence is as follows:

$\forall x[B(x) \rightarrow \forall y(F(y) \rightarrow \neg L(x,y))]$

$\Leftrightarrow \forall x \forall y [B(x) \rightarrow (F(y) \rightarrow \neg L(x,y))]$ by null quantification law

$\Leftrightarrow \forall x \forall y [(B(x) \wedge F(y)) \rightarrow \neg L(x,y)]$ by exportation rule

$\Leftrightarrow \forall x \forall y [(F(y) \wedge B(x)) \rightarrow \neg L(x,y)]$ by commutative rule

$\Leftrightarrow \forall x \forall y [F(y) \rightarrow (B(x) \rightarrow \neg L(x,y))]$ by exportation rule

If the quantifiers of a propositional function are either all universal or all existential, then their order is irrelevant. Hence,

$\Leftrightarrow \forall y \forall x [F(y) \rightarrow (B(x) \rightarrow \neg L(x,y))]$

$\Leftrightarrow \forall y [F(y) \rightarrow \forall x(B(x) \rightarrow \neg L(x,y))]$ by null quantification

Since the choice of $x$ or $y$ is arbitrary, we can swap them as long as we are consistent. Hence,

$\Leftrightarrow \forall x [F(x) \rightarrow \forall y(B(y) \rightarrow \neg L(y,x))]$ by null quantification

Therefore,

$\forall x[B(x) \rightarrow \forall y(F(y) \rightarrow \neg L(x,y))] \Leftrightarrow \forall x [F(x) \rightarrow \forall y(B(y) \rightarrow \neg L(y,x))]$

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I would like to make a small correction to the natural interpretation of your statements, however. The first statement $\forall x[B(x) \rightarrow \forall y(F(y) \rightarrow \neg L(x,y))]$ reads as:

"For all things x, if x is a bee, then for all things y, if y is a flower, then x hates y."

The second statement $\forall x [F(x) \rightarrow \forall y(B(y) \rightarrow \neg L(y,x))]$ reads as:

"For all things x, if x is a flower, then for all things y, if y is a bee, then y hates x."