# Math Help - Monadic Predicate Logic - Hardest Problem in Problem Set

1. ## Monadic Predicate Logic - Hardest Problem in Problem Set

Symbolize the following: Given that some mean elf will bite and some friendly one will too, the mean ones will bite whether or not provoked but the friendly ones will bite only if provoked.

Attempt:
$(\exists x((Gx \wedge Hx) \wedge Ix) \wedge \exists x((Fx \wedge Hx) \wedge Ix)) \rightarrow \forall x ((((Gx \wedge Hx) \wedge Ix) \rightarrow (Jx \vee \neg Jx)) \wedge ((Fx \wedge Hx) \wedge Ix) \rightarrow Jx)$

Symbolization Key:
F{1} : {1} is friendly
G{1} : {1} is mean
H{1} : {1} is an elf
I{1} : {1} will bite
J{1} : {1} is provoked

This is the last and unfortunately the hardest question in my problem set. I think my error lies within the second part of my attempt, though I'm not sure.

2. ## Re: Monadic Predicate Logic - Hardest Problem in Problem Set

lets try to analyse it one by one, first part seems ok. now consider the sentence,

"the mean ones will bite whether or not provoked" is equivalent to

"so whenever we have an elf which is mean it will bite"..... which means

"if the x is an elf and x is mean then x will bite"........

you will have to apply universal quantifier to above statement.....now the word "but" tells us that we have a conjunction with the above statement. also notice the phrase "only if" in the last part of the sentence. so we can convert that sentence as

"if the x is an elf and x is friendly and x bites then x is provoked"

you will have to apply universal quantifier here as well.........

so in the second part you have a conjunction and both the conjuncts have universal quantifier to it. since universal quantifier distributes over conjunction , you can use the universal quantifier at the beginning of the second part.

so can you convert the second part into the symbols ?

3. ## Re: Monadic Predicate Logic - Hardest Problem in Problem Set

I got it, thanks!