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+18 votes

Given that

  • $B(x)$ means "$x$ is a bat",
  • $F(x)$ means "$x$ is a fly", and
  • $E(x, y)$ means "x eats $y$",

what is the best English translation of $$ \forall x(F(x) \rightarrow \forall y (E(y, x) \rightarrow B(y)))?$$

  1. all flies eat bats
  2. every fly is eaten by some bat
  3. bats eat only flies
  4. every bat eats flies
  5. only bats eat flies
asked in Mathematical Logic by Veteran (97.7k points) | 819 views
Whosoever will eat fly will become bat. Hehe :)
just want to know is this mean

∀x(F(x)→∀y(E(y,x)→B(y)))? every fly is eaten by every bat
For all x if its a fly then for all y if y eats x then y is a bat ... option E ..

3 Answers

+27 votes
Best answer
If $x$ is a fly, then for all $y$ which eats $x$, $y$ is a bat. This means only bats eat flies. Option (E).
answered by Veteran (414k points)
edited by
how d) is wrong ?
There can be a bat which does not eat a fly.
ok got it ,  so for d , it  will be converse means b(y) imples e(x,y) rt ?
Thanks :)

$\text{All flies eats bats}$ 

$\forall x (F(x) \rightarrow \exists y (E(x,y) \wedge B(y)) ) \text{As it is not given that all flies eats all Bats}$

$\text{Every fly is eaten by some bat}$

$\forall x (F(x) \rightarrow \exists y (E(y,x) \wedge B(y)))$

$\text{bats eat only flies}$

$\forall x (B(x)\rightarrow \exists y (E(x,y) \wedge F(y)))$

$\text{every bat eats flies}$

$\forall x(B(x)\rightarrow \exists y(E(x,y) \wedge F(y)))$

@Arjun sir can you verify all , i am not sure about option$3$

yes, third is not correct because some $y$ such that $x$ does not eat $y$ is enough to make the whole expression TRUE.
sir now is it correct (i have updated commnet)?
No, it now means every bat eats some fly.
Sir one more try.

$\forall x \exists y ((B(x) \wedge E(x,y))\rightarrow F(y))$
Why you use exists for $y$ and not forall?
$\forall x \forall y ((B(x) \wedge E(x,y))\rightarrow F(y))$

Is it okk ? if not please tell the answer :D
How can we say that "only" ?
Usually, simplification of such predicate calculus makes it a lot easier to understand.

$$\begin{align}&\forall x F(x) \to \forall y(\neg E(y,x) \lor B(y)) \\ & \equiv \forall x \left[ \neg F(x) \lor  \forall y(\neg E(y,x) \lor B(y)) \right] \\ & \equiv \forall x \forall y \left[ \neg F(x) \lor \neg E(y,x) \lor B(y) \right]\end{align}$$
Use De Morgan's law
$$\begin{align}& \forall x \forall y \left[ \neg (F(x) \land E(y,x)) \lor B(y) \right] \\ & \equiv  \forall x \forall y (F(x) \land E(y,x)) \to B(y) \end{align}$$
Now it is easier to interpret it as option E
Sir, isn't option (b) and (e) equivalent to each other? Why option (B) is not the correct choice?
+9 votes
For any/all x if it is a fly and if it gets eaten up by any y than that y must be a bat..

So only bats eat flies.

Option e
answered by (111 points)
+1 vote

first solve ∀y(E(y,x)→B(y)) only take x=fly1

so for x1fly1  ∀y(E(y,x)→B(y)) telling that : (if y1 eats fly1  then y1 is Bat) and(if y2 eats fly1  then y2 is Bat) and (if y1 eats fly1  then y1 is Bat ) and ....(if yn eats fly1  then yn is Bat).

this conclude that:  only bats eats fly1 .

no solve comlete :∀x(F(x)→∀y(E(y,x)→B(y)))

 = (if x is fly1 then only bats eats fly1) and (if x is fly2 then only bats eats fly2) and (if x is fly3 then only bats eats fly3)...

= every fly  eaten by bats only.

= only bats eats flies.

Ans: E


answered by Active (3.4k points)

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