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
