Difficult Predicate Logic Symbolization

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The discussion focuses on symbolizing a complex sentence involving mean and friendly elves and their biting behavior based on provocation. The initial attempt at symbolization is critiqued for including unnecessary components, such as the always-true disjunction regarding provocation. Suggestions are made to simplify the notation and clarify the relationships between the variables, specifically questioning the need for separate variables for mean and friendly elves. The conversation emphasizes the importance of clear notation and logical structure in predicate logic symbolization. Participants continue to seek clarity on how to effectively represent the conditions of biting behavior in logical terms.
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Homework Statement



I do not know if this is the right forum or subforum for this kind of topic. So if it is not, I apologize in advance.Symbolize the following sentence:

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.

Homework Equations



Fx: x is friendly
Gx: x is mean
Hx: x is an elf
Ix: x will bite
Jx: x is provoked

The Attempt at a Solution

This is my attempt and to me it seems right, but I could be missing one tiny simple thing.

∃x∃y(((Hx∧Gx∧Ix)∧(Fy∧Hy∧Iy))→((Jx∨∼Jx)→Ix)∧(Iy→Jy))
 
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Ok, given I'm actually not that good at these...

First for clarity and to simplify reading, I would probably use the folliwng nonmenclature (some good notation can simplify/clarify your problem immensly)
E-Elf, M-Mean, F-Friendly, P-Provoked, B-Bites

if I look at yours
∃x∃y(((Ex∧Mx∧Bx)∧(Fy∧Ey∧By))→((Px∨∼Px)→Bx)∧(By→Py) )

first I note (Px∨∼Px) is always true, as the elf is either provoked or not, so I'm not sure you need that? so it becomes
∃x∃y(((Ex∧Mx∧Bx)∧(Fy∧Ey∧By))→(Bx)∧(By→Py)

then I'd say both Bx and By are assumed true in the LHS, so are always true when the RHS is evaluated? so it becomes
∃x∃y(((Ex∧Mx∧Bx)∧(Fy∧Ey∧By))→Py

Which doesn't tell us a heap... and then I would ask if you really need the x and the y to cover the two cases when the elf is mean or friendly, and whether friendly and mean were mutually exclusive so you could you assume friendly = (not mean), and the "for exists" only shows its true for some x and y, not all of them...

So I was thinking you could simplify your approach by dropping the x and ys, and looking at it as follows...

Given that some mean elf will bite and some friendly one will too,
(is an elve and bites)
the mean ones will bite whether or not provoked but the friendly ones will bite only if provoked.
(implies either mean or friendly and provoked)
 
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Thanks for your time lanedance, but I am still stuck!
 
ok how about looking at the first bracketed comment, how would you write that?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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