# How to solve this type of Puzzles (Syllogism)?

I have seen lot of questions of below type predictions. I can't figure out the answer. Is there any common method to solve this ?

Statements:-

All cats are dogs.
All dogs are cows.
Some dogs are pigs.
Some chicken are dogs.

Q: Say true or false

A. Some chicken are pigs.
B. All pigs are cats.
C. Some cows are chicken.


What is the answer and please explain how you solved this ?

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These are are called syllogism, there are various techniques that one could use while solving them the Wikipedia page has a decent discussion.

• A. Some chicken are pigs. (False)
• B. All pigs are cats. (False)
• C. Some cows are chicken. (True)
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Saying "false" for A and B is probably what the problem wants, but actually they are undecidable from the axioms given -- that is, it is possible but not necessary that they are true. For example, in a world containing a single animal that is simultaneously cat, dog, cow, pig and chicken, all seven statements would be true. –  Henning Makholm Dec 9 '11 at 12:48
@Henning:What I understand about syllogisms is that to be valid the conclusion should be logically derived from the premises else the syllogism is invalid.I guess here, False means invalid. –  Quixotic Dec 9 '11 at 15:30
Modal logic addresses this distinction. –  Dan Brumleve Jan 24 '12 at 5:52

Let $C$ be the set of all cats; $D$ the set of all dogs; $P$ the set of all pigs; $H$ the set of all chickens (for "hens"); $W$ the set of all cows.

Statement 1 says that $C\subseteq D$.

Statement 2 says that $D\subseteq W$.

Statement 3 says that $C\cap P\neq\varnothing$.

Statement 4 says that $D\cap H\neq\varnothing$.

Statement A says $H\cap P\neq\varnothing$. This cannot be deduced from Statements 1-4; for example, say $C=\{a,b\}$, $D=\{a,b,c\}$, $W=\{a,b,c,d\}$, $P=\{b,q\}$, $H=\{c\}$. Then Statements 1-4 are true, but $H\cap P=\varnothing$.

Statement B says that $P\subseteq C$; again, the example above shows that this cannot be deduced from Statements 1-4 alone.

Statement C says that $W\cap H\neq\varnothing$. This is true: we know there exists some $x\in D\cap H$ (by statement 4), and that $D\subseteq W$ (by statement 2); hence $x\in W$. Since $x\in H$, we obtain $x\in W\cap H$, which is therefore nonempty.

So the syllogisms

Statement 1
Statement 2
Statement 3
Statement 4
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Statement X


are invalid if X is A or B, and is valid when X is C.

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