Question 1042040
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I'm going to use the convention that upper case refers to a name and lower case refers to an occupation and the symbol <> means 'not equal'.

The four statements are then:

B = s
C = b
D <> s
S <> d

Let us assume that the first statement is true, so the other three are false.
Given that the four True statements are:

B = s
C <> b
D = s
S = d

But if C <> b, and because C = c is not allowed by the problem statement, C = d or C = s.  Either one of these statements contradicts D = s or S = d.  Therefore the assumption that B = s is false.

Next, let us assume that the second statement is true, and the other three are false.

The four true statements are then:

B <> s
C = b
D = s
S = d

Since B <> s, it must be true that B = c or B = d, hence Baker is the carpenter because B = d contradicts S = d.

Next, let us assume that the third statement is true, leading to:

B <> s
C <> b
D <> s
S = d

B <> s means B = c or B = d.  C <> b means C = s or C = d.  And D <> s means D = b or D = c.  B must equal c because B = d contradicts S = d.  Then D = b and C = s.  Again, Baker is the carpenter.

Lastly, let us assume that the fourth statement is true, leading to:

B <> s
C <> b
D = s
S <> d

Since D = s and C must then be d or s, C = d.  S <> d means S = b or S = c and B <> s means B = c or B = d, so S = b and B = c.  Again, Baker is the carpenter.

We don't know which of statements 2, 3, or 4 is true, but we do know for certain that Baker is the carpenter.

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John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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