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Which one of the first order predicate calculus statements given below correctly expresses the following English statement?

Tigers and lions attack if they are hungry or threatened.

1. $∀x[(\text{tiger}(x) ∧ \text{lion}(x)) → {(\text{hungry}(x) ∨ \text{threatened}(x)) → \text{attacks}(x)}]$
2. $∀x[(\text{tiger}(x) ∨ \text{lion}(x)) → {(\text{hungry}(x) ∨ \text{threatened}(x)) ∧ \text{attacks}(x)}]$
3. $∀x[(\text{tiger}(x) ∨ \text{lion}(x)) → {\text{attacks}(x) → (\text{hungry}(x) ∨ \text{threatened}(x))}]$
4. $∀x[(\text{tiger}(x) ∨ \text{lion}(x)) → {(\text{hungry}(x) ∨ \text{threatened}(x)) → \text{attacks}(x)}]$
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The statement $\text{Tigers and lions attack if they are hungry or threatened"}$ means that if an animal is either tiger or lion, then if it is hungry or threatened, it will attack. So option (D) is correct.

Do not get confused by $\text{"and"}$ between tigers and lions in the statement. This $\text{and"}$ does not mean that we will write $\text{"tiger(x) ∧ lion(x)"}$, because that would have meant that an animal is both tiger and lion, which is not what we want.

http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2006.html

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can you please explain the difference between (C) and (D)

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It has been said that tiger and lion attack if they are hungry or threatened.
It means if they are hungry or threatened, they will attack. But if they are not, they may or may not attack(nothing is said about this). In option (C) , it is attack(x)->(hungry(x) or threatened(x)) , this is not true , as if attack(x) is true, the 2nd part has to be true, i.e. they HAVE to be hungry or threatened, which is not correct. Hence D is the correct choice
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Thanks dude, that clears it.
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∀x [ ( Tiger(x) ∨ Lion(x) ) ∧  (Hungry(x) ∨ Threatened(x)) --> Attack(x) ]

We can represent also in this way,bcoz of

Exportation law : (P-->(Q-->R)) ≡ ((P ∧ Q) -->R)

+1
I am unable to differentiate between option B and D :(
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@♥_Less  Option B is false cause

∀x[(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) ∧ attacks(x)}]

This statement can turn false in one case where :

(tiger(x) ∨ lion(x)) is True

(hungry(x) ∨ threatened(x)) is false

attacks(x) is True

getting one such case can turn the entire statement false. Eventhough there exist a case where :-

(tiger(x) ∨ lion(x)) is True

(hungry(x) ∨ threatened(x)) is True

attacks(x) is True

cause "for all"  ∀ (x) quantifier is a collection of conjunction it should ignore this case :-

(tiger(x) ∨ lion(x)) is True

(hungry(x) ∨ threatened(x)) is false

attacks(x) is True

by making it True

This is the flaw of option B The statement says Tiger or lion will attack if it is hungry or threatened

but it does not says It will not attack if it is not (hungry or  threatened ),{It may or may not attack that is not our concern ,Our matter of concern will be when It is hungry or threatened and still it doesn't attack only in this case the statement should turn false }

Now place the same case on option D it will not turn false,it will simply ignore such cases by making it True .That is why Option D is the only answer .

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B) ∀x[(tiger(x)∨lion(x))→(hungry(x)∨threatened(x))∧attacks(x)]

D) ∀x[(tiger(x)∨lion(x))→(hungry(x)∨threatened(x))→attacks(x)]

The first part upto this (tiger(x)∨lion(x)) is same. Main problem lies in finding the difference b/w the 2nd part.

If you are finding it difficult then create a Truth Table for it though it may take time. I tried doing this to show the clear difference.

I am considering only one animal "x" which is a tiger so tiger(x)∨lion(x) = T+F=T.

For the 2nd part :

For B)

 hungry(x) threatened(x) attacks(x) {hungry(x)∨threatened(x)} ∧ attacks(x) T T T T T T F F T F F F T F T T F T T T F F T F F T F F F F F F

For D)

 hungry(x) threatened(x) attacks(x) {hungry(x)∨threatened(x)} -> attacks(x) T T T T T T F F T F F F T F T T F T T T F F T T F T F F F F F T

I have marked the difference in BLUE. After creating this table it will be easy to analyze what made the difference. Check the set of inputs i.e. hungry(x),threatened(x) and attacks(x).

In the first table we can see that if tigers or lions are not hungry and still they attack (attack(x)=T) then outcome is false. If they do not attack (attack(x)=F) then also outcome is false. But this shouldn't be the case because in the question it is given that they if they are hungry or threatened then they will attack but nothing is mentioned about what action will they take if they are not hungry or threatened(hungry(x)=F and threatened(x)=F). So whatever action they take (whether attack(x)=T or attack(x)=F), it shouldn't turn the clause to False.

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Why option B is wrong
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Also, q if p type of clause in the statement given makes option (C) incorrect.
For solving these type of questions break the sentence into predicates i.e. p(x)=lions or(not and) tigers q(x)=hungry or threatened r(x) = attack now join them using implication for all x so option d.

∀x [ ( Tiger(x) ∨ Lion(x) ) ∧  (Hungry(x) ∨ Threatened(x)) --> Attack(x) ]

We can represent also in this way,bcoz of

Exportation law : (P-->(Q-->R)) ≡ ((P ∧ Q) -->R)

So, D is correct ans.

(D)  ∀x[(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}]

option D
+1 vote

Option D bcz it may be lion or tiger here which attacks and AND is just used as 'OR' option a says they always attack together which is wrong so not option a

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Here "AND" means an animal is both tiger and lion, which is not what we want.
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tiger and lion attack if they are hungry or threatened.
This can be written as

(tiger and lion ).   (attack )if (they are hungry or threatened.)

(anything in the world if is is tiger v lion ) (attack ) if ( hungry or threatened)

(attack ) if ( hungry or threatened) is q if p form hence p->q

if ( hungry or threatened)->(attack) rest is simple

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