Problem Statement: Tony, Shi-Kuo and Ellen belong to the Hoofers Club. Every member of the Hoofers Club is either a skier or a mountain climber or both. No mountain climber likes rain, and all skiers like snow. Ellen dislikes whatever Tony likes and likes whatever Tony dislikes. Tony likes rain and snow. Query: Is there a member of the Hoofers Club who is a mountain climber but not a skier?

Convert it into first order logic.

M(x) : x is member of Hoofer Club.

S(x) : x is a Skier

C(x) : x is a Mountain climber

L(x,y) : x likes y

where

x is a member of Hoofer Club

y is either "snow", "rain"

KNOWN STATEMENTS

1. M(x) => S(x) V C(x)
2. C(x) => ~L(x,Rain)
3. S(x) => L(x,Snow)
4. L(Ellen,y) <==> ~L(Tony, y)
5. L(Tony, Rain)
6. L(Tony, Snow)

Consider Statement 4, 5 and 6

L(Ellen,y) <==> ~L(Tony, y) AND L(Tony, Rain) AND L(Tony, Snow)

contrapositive of statement 4

L(Tony, y) <==>~L(Ellen,y)

==> 7. ~L(Ellen,Rain)

==>8. ~L(Ellen,Snow)

By modus ponen

Consider Statement 3 and 8

S(x) => L(x,Snow) and ~L(Ellen,Snow)

Taking contrapositive

~L(x,Snow) ==> ~S(x)

=>9. ~S(Ellen) // Ellen is not a skier

By modus ponen

Consider statements 2 and 5

C(x) => ~L(x,Rain)

Taking Contrapositives

L(x,Rain) => ~C(x)

and

L(Tony, Rain)

=> 10. ~C(Tony) // Tony Not A Mountain Climber

By modus ponen

Consider statements 1 and 9

M(Ellen) => S(Ellen) V C(Ellen) and ~S(Ellen)

and

M(Ellen)

=> S(Ellen) V C(Ellen) and ~S(Ellen)

=> C(Ellen) // Ellen is a Mountain Climber but not skier

Consider statements 1 and 10

M(x) => S(x) V C(x) and ~C(Tony)

and

M(Tony)

=> C(Ellen) // Ellen is a skier but not mountain climber

ANSWER : ELLEN

1.How to turn this logic into prolog and.

2. Write the prolog code.