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From: Gadfly on 8 Aug 2010 19:32
Hi All --

I'm a newbie asking for help translating an English sentence to first-
order logic. The sentence is "The only thing with nothing to its
right is the medium cube," and is meant to be true according to the
Russellian analysis of definite descriptions.

This sentence appears as an exercise in Tarski's World. Most of you
are probably familiar with the software / text. For those of you who
aren't, it allows users to design 3 dimensional worlds made of objects
of various shapes and sizes, and write sentences in FOL which describe
elements of that world. It also comes with a grading program called
the "Grade Grinder" which evaluates the worlds and / or sentences
submitted by the user for correctness with respect to a given exercise
in the text. Most of the time, the "Grade Grinder" is able to tell
the student if their worlds or sentence are either correct or
incorrect. On rare occasions, it is unable to determine the
correctness or incorrectness of the submission, and basically tells
the user to go check with his / her instructor. This is situation I'm
in now.

I'm doing Tarksi's on my own just to learn about logic (Scout's
honor!), and have no instructor to turn to for help, so I thought
someone belonging to this group might be able to tell me if my
translation is correct or not. Here it is:

"The only thing with nothing to its right is the medium cube."

∃x (Cube(x) ∧ Medium(x) ∧ ∀y ((Cube(y) ∧ Medium(y)) → (x = y)) ∧ ¬∃y
RightOf(y, x) ∧ ∀y ((y ≠ x) → ∃z RightOf(z, y)))

Here's how I broke it down.

A medium cube = ∃x (Cube(x) ∧ Medium(x))

The medium cube = ∃x (Cube(x) ∧ Medium(x) ∧ ∀y ((Cube(y) ∧ Medium(y))
→ (x = y))

Something with nothing it its right = ∃x ¬∃y RightOf(y,x)

The medium cube with nothing to its right = ∃x (Cube(x) ∧ Medium(x) ∧
∀y ((Cube(y) ∧ Medium(y)) → (x = y)) ∧ ¬∃y RightOf(y, x)

I translated the condition "only" by saying "all objects that are not
the medium cube have something to their right:" ∀y ((y ≠ x) → ∃z
RightOf(z, y)))

Put all together, the final sentence is
∃x (Cube(x) ∧ Medium(x) ∧ ∀y ((Cube(y) ∧ Medium(y)) → (x = y)) ∧ ¬∃y
RightOf(y, x) ∧ ∀y ((y ≠ x) → ∃z RightOf(z, y)))

But when I submit it to the Grade Grinder, it barfs back this message:

Sentences 3.35 (Student file: "Sentences 3.35.sen")
We found problems in your sentences:
??? We could not determine whether your ninth sentence, "Ex
(Cube(x) &
Medium(x) & Ay ((Cube(y) & Medium(y)) -> (x = y)) & ~Ey
RightOf(y, x)
& Ay ((y # x) -> Ez RightOf(z, y)))", was correct: The Grade
Grinder
timed out while checking your answer. Your instructor will
have to
verify this answer.

Can anyone help?

Many thanks,
Dave Yrueta




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