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

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

false

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8

step-by-step explanation:

to find the amount of gift bags to use all the supplies, find the gcf or greatest common factor. this is the largest number which multiplies evenly into each term.

48 has factors: 1, 2, 3, 4, 6, 8, 12. 16, 24, 48

80 has factors: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80

64 has factors: 1, 2, 4, 8, 8, 16, 32, 64

the highest number factor they all share in common is 8. this means there will be 8 bags with 6 pencils, 10 markers, and 8 stickers in each bag.