Question

Consider the following information: Tony, Mike, and John belong to the Alpine Club. Every member of the Alpine club who is not a skier is a mountain climber. Mountain climbers do not like rain, and anyone who does not like snow is not a skier. Mike dislikes whatever Tonty likes, and he likes whatever Tony dislikes Tony like rain and snow.

Represent the information as a set of FOPL statements appropriate for backward chaining. Show how an answer io he qury "s iheore a rmernber of the Alpine club who is a mouniain climber but not a skier2" is found by a backward chaining automated deduction system by providing a (consistent) AND/OR solution tree that solves and answer. Clearly show the rules and facts separately, and number them.

Some (part ofl a Suggested Ontology:

Member(x) x is member of Alpine club
Skier(y) y is a skier
Climber(z) z is a mountain climber
Likes(x,y) x likes y

Answer 1

Answer:

ranslation into first order logic ,

Tony, Mike and John belong to Alpine club.

S1 Member (Tony)

S2 Member (mike)

S3 Member (john)

Every member of the Alpine club who is not a skier is a mountain climber

S4 \forallx(Member(x)\wedge~Skier(x)\supsetClimber(x))

Mountain climbers do not like rain

S5 \forallx(Climber(x) \supset ~Like(x,Rain))

Anyone who does not like snow is not a skier

S6 \forallx(~Like(x,snow) \supset ~ Skier(x))

Mike dislikes whatever Tony likes

S7 \forallx(Like(Tony,x) \supset ~ Like(mike,x))

And likes whatever Tony dislikes

S8 \forallx(~Like(Tony,x) \supset Like(Mike,x)

Tony likes rain and snow

S9 Like(Tony,rain)

S10 Like(Tony, snow)

From s10 we know that (I(tony),I(snow)) \in I(Like)

From s7 we know that for every assignment v

(D,I),v|= Like(tony,x)\supset ~Like(Mike,x)

(D,I),v|= Member(x) \wedge Climber(x) \wedge ~ Skier(x)

So

(D,I),v |= \existsx(Member(x)\wedgeClimber(x)\wedge~Skier(x))

Hence a member of Alpine club who is a mountain climber but not a skier

suppose we donot have S7 , we have only s1-s6 and s8-s10.

To prove , we have to produce interpretations as :

D ={ t,m,j,s,r }

Interpretations:

I(tony)=t, I(mike)=m, I(john)=j, I(snow)=s, I(rain)=r

I(member)= {t,m,j}

I(skier)= {t,m,j}

I(climber)= {}

I(Like)= {(t,s),(t,r),(m,s),(m,r),(m,m),(m,t),(m,j),(j,s)}

Hence a member of Alpine club who is a mountain climber but not a skier