MTH110 – KNIGHTS AND KNAVES
EXPLANATIONS
1. All the knights and knaves problems assume that each person encountered is either a knight or
a knave, but not both. More formally, if we decide to represent the statement “P is a knight”
by p, then “P is a knave” is represented by ~p
2. All knights and knaves problems assume that knights always tell the truth and knaves always
lie. This means that each utterance any of the character makes is formalized into 2 statements:
“if the character is a knight then the utterance is true” and “if the character is a knave, then the
utterance is false”.
3. In addition to the rules of inference in table 1.3.1, you can assume that:
(a b)  a  b
(Equivalence of therefore and implication)
This is because a b is an alternate notation for a  b (i.e. a  b is a tautology).
So since a b is really stating that a  b is always true, if your proof derives that
a b, then you can conclude that a  b is true.
This is very useful for knights and knaves problems because these problems are usually solved
using the rule of contradiction: p  c ~p.
In practice you usually make the assumption that one of the characters is a knight (or a knave),
and after a chain of deductions, you encounter a contradiction, which you then use to deduce
that your initial assumption was false.
MTH110 – KNIGHTS AND KNAVES
EXAMPLE IN BOOK
Problem:
A says: “B is a knight”
B says: “A and I are of opposite types”
What are A and B
Parametrisation of problem:
Let a = A is a knight
Let b = B is a knight
A is saying: b
i.e. we know:
B is saying: a  ~b i.e. we know:
Reasonning
Assume: a (5)
 b
(6)
 (a  ~b)
 (a  ~b)  (~b  a)
 a  ~b
 ~b
 b  ~b
c
ac
 ~a
(7)
 ~b
(8)
(1): a  b
and
(3): b  (a  ~b) and
(2): ~a  ~b
(4): ~b  ~(a  ~b)
by (1) and Modus Ponens
by (3) and Modus Ponens
by definition of 
by specialization
by (1) and Modus Ponens
by (6) and conjunction
by negation law
by (5) and equivalence of therefore and implies
by rule of contradiction
by (2) and Modus Ponens
Solution: The solution is in (7) and (8): A and B are both knaves.
Verification
You can verify that this problem is not a paradox by reasoning with the starting assumption that a
is a knave. This is left as an exercise.