TIFR CSE 2017 | Part B | Question: 11

Question

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

• A. all flies eat bats
• B. every fly is eaten by some bat
• C. bats eat only flies
• D. every bat eats flies
• E. only bats eat flies

Comments

For all x if its a fly then for all y if y eats x then y is a bat ... option E ..

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Interesting fact is this question repeated in ISRO 2020 (only bears eat fish)

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1. All flies eat bats:
∀x (F(x) → ∃y (Eats(x, y) ∧ B(y)))

 2. Every fly is eaten by some bat:
∀x (F(x) → ∃y (Eats(y, x) ∧ B(y)))

 3. Bats eat only flies:
∀x (B(x) → ∀y (Eats(x, y) → F(y)))
or
∀x∀y (B(x)∧Eats(x, y)→F(y))

 4. Every bat eats flies:
∀x (B(x) → ∃y (Eats(x, y) ∧ F(y)))

 5. Only bats eat flies:
∀x∀y (F(x) ∧ Eats(y, x)  → B(y))
or
∀x (F(x)→∀y (Eats(y, x) → B(y)))

Answer 1

If $x$ is a fly, then for all $y$ which eats $x$, $y$ is a bat. This means only bats eat flies.

Option (E).

Comments

how d) is wrong ?

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There can be a bat which does not eat a fly.

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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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$\text{All flies eats bats}$ 

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

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

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

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$\text{bats eat only flies}$

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

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$\text{every bat eats flies}$

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

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

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

Is it okk ? if not please tell the answer :D

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How can we say that "only" ?

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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

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Sir, isn't option (b) and (e) equivalent to each other? Why option (B) is not the correct choice?

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@Arjun sir, @sourav.

Is this correct for option c

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@Arjun sir But if y eats x then only it will be implied as a bat??so how you can say that not all bats eat fly?please explain

Answer 2

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

Answer 3

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