M325K
HW 02  Solutions to Graded Questions
(
Out of 25 points)
Section 1.3
23.
(4pts) Its form:
q
p
∨
q
p
→
r
q
~
∨
∴
⎯
⎯
⎯
⎯
⎯
⎯
→
←
premises
⎯
⎯
⎯
→
←
conclusion
28
. (2pts) Its form is:
q
p
→
Invalid, a converse error.
q
p
∴
38.
(5pts) c) Answer: There is only one knave and only one knight.
There are 4 cases to consider, namely
a) Both are knights
c) E is a knight and F is a knave.
b) Both are knaves
d) E is a knave and F is s knight.
Observe that (a) is not possible, because then both would also be telling the truth and both would
also be knaves. Similarly, (b) is not possible because in this case both would be telling the truth
which would make them knights, a contradiction. Hence there is only one knight and only one
knave. Note that in case c and d, both E and F would have spoken as reported, without a
contradiction.
It is OK if you combine c and d and consider it as
“if one is a knight and the other is a knave”. But
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 Spring '10
 seckin
 Logic, muscles, Modus ponens, Modus tollens, Socko

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