903 views

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
| 903 views
+5
Whosoever will eat fly will become bat. Hehe :)
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just want to know is this mean

∀x(F(x)→∀y(E(y,x)→B(y)))? every fly is eaten by every bat
+2
For all x if its a fly then for all y if y eats x then y is a bat ... option E ..

If $x$ is a fly, then for all $y$ which eats $x$, $y$ is a bat. This means only bats eat flies. Option (E).
by Veteran (423k points)
edited by
0
how d) is wrong ?
+2
There can be a bat which does not eat a fly.
+2
ok got it ,  so for d , it  will be converse means b(y) imples e(x,y) rt ?
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Exactly.
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Thanks :)
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$\text{All flies eats bats}$

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

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

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

$\text{bats eat only flies}$

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

$\text{every bat eats flies}$

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

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

+1
yes, third is not correct because some $y$ such that $x$ does not eat $y$ is enough to make the whole expression TRUE.
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sir now is it correct (i have updated commnet)?
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No, it now means every bat eats some fly.
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Sir one more try.

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

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How can we say that "only" ?
+5
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
0
Sir, isn't option (b) and (e) equivalent to each other? Why option (B) is not the correct choice?
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
by (111 points)

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

by Active (3.6k points)