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Summary: Brachman and Levesque, chapter 2 exercise 4 .
Question .
Formulate the requirements below as sentences of first order logic and show
that the two of them cannot be true together in any interpretation. (This is the
barber's paradox by Bertrand Russell.)
1. Anyone who does not shave himself must be shaved by the barber [note
the barber - some unique person, not a barber].
2. Whomever the barber shaves, must not shave himself.
Hints: introduce a constant for the barber and a binary predicate Shaves(x, y).
Answer .
Translation:
1. x(ŽShaves(x, x) Shaves(barber, x))
2. x(Shaves(barber, x) ŽShaves(x, x))
Suppose some interpretation (D, I) satisfies both 1 and 2. There is an object
in D which is the meaning of `barber' in that interpretation, I(barber). Let us
call it b. This b either shaves himself or not, or in logic speak the pair (b, b) is
either in the interpretation I(Shaves) of the predicate Shaves, or not.
First suppose it is: (b, b) I(Shaves). Then since the second sentence is
true in (D, I), for every assignment v:
(D, I), v |= Shaves(barber, x) ŽShaves(x, x)
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