Homework 1
February 13, 2013
This assignment is due in two weeks, i.e. by Thursday 28 February, say
at midnight. You may either leave your answers as hardcopies in Shengyang
Zhong’s mailbox at ILLC, or else send them electronically at his email address [email protected], with CC to me (at [email protected]).
Please type in capital letters on the first page of your answers the
following: your full name (both first name and family name).
1. Alice has a natural number na ∈ N = {0, 1, 2, . . .} written on her
forehead and Bob has a natural number nb ∈ N = {0, 1, 2, . . .} written
on his forehead. It is common knowledge that (a) each of them can
see the other’s number, but neither of them can see his/her number;
(b) one of the two numbers is the immediate successor of the other
(i.e. either na = nb + 1 or nb = na + 1); (c) the two children (Alice
and Bob) are perfect logicians.
The Father asks them, repeatedly: “Do you know your own number?”.
The two are supposed to answer truthfully, publicly, simultaneously
and independently (without any other communication). they both
answer “I don’t know” to the first 17 questions, after which Alice
answers “Yes, now I know my number” to the 18th question (while
Bob still answers “I don’t know”).
• What is Bob’s number?
• Will Bob ever know his number? If so, when will he answer “I
know my number”?
• Prove your conclusions semantically, by drawing or representing mathematically the initial epistemic model, then applying
repeated updates with the children’s answers.
NOTE: Since the model is infinite, you may want to make your drawing
precise by adding a mathematical description.
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2. Alice and Bob are each given a card with a number written on it,
say nA is Alice’s number and nB is Bob’s number. The following
facts are common knowledge: (1) nA and nB are natural numbers, i.e.
nA , nB ∈ {0, 1, 2, 3, . . .}; (2) Alice’s number is either equal to Bob’s or
is the immediate successor of Bob’s, i.e. nA − nB ∈ {0, 1}; (3) each
of the two can see the number written on his/her card, but not the
other’s number; (4) Alice and Bob are perfect logicians and they never
lie.
The following dialogue is observed. First, Alice tells Bob: “I don’t
know your (i.e. Bob’s) number”. Then, Bob answers to Alice: “I
don’t know your number either”. Next, Alice tells Bob: “I still don’t
know your number”. But then Bob answers: “ Oh, but now I do know
your number!”. Alice replies: “Well, but now I know your number as
well”.
What are the numbers? Solve this puzzle semantically, by drawing
the initial state model and performing the successive updates with the
sentences announced by Alice and Bob.
3. Solve the Muddy Children puzzle in general (n children, k of whom
are dirty) proof-theoretically, by proving the sentence
(exactly k∧C2vision) ⇒ [!(at least one)][!(nobody knows)]k−1 dirties know
using the axioms and rules of P AC. (Here, we use all the abbreviations
introduced in the slides of Lecture Notes Week 2.) In your proof you
might skip all the deductive steps involving only purely propositional
reasoning (i.e. the rules and axioms of Propositional Calculus), but do
try to justify all the other deductions. One possible proof may involve
the following steps:
(a) First, using the PAC axiom
[!φ](ψ ⇒ θ) ⇔ ([!φ]ψ ⇒ [!φ]θ)
and the Reduction Axioms, prove that: if ψ is either an atomic
sentence p or its negation ¬p, then we have
[!φ](ψ ⇒ θ) ⇔ (ψ ⇒ [!φ]θ)
(b) Prove that
(exactly k+1 ∧ C2vision) ⇒ (da ⇒ 2a (¬da ⇒ (exactly k ∧ C2vision)))
(c) In the following, I use the abbreviation
≥1
:=
2
at least one.
Now go back to the sentence encoding the Muddy Children scenarion and prove its initial case (k = 1), i.e. show that:
(exactly 1 ∧ C2vision) ⇒ [!(≥ 1)]dirties know
(d) Using the formulas proved in the last three items, prove the inductive step of the Muddy Children scenario: i.e. assuming as
proven (the case k) that
(exactly k ∧ C2vision) ⇒ [!(≥ 1)][!(nobody knows)]k−1 dirties know ,
use this assumption together with (c) to show that
(exactly k+1∧C2vision) ⇒ (da ⇒ 2a (¬da ⇒ [!(≥ 1)][!(nobody knows)]k−1 dirties know)),
Then (using one of the two implications forming the ActionKnowledge Axiom) show that
(exactly k+1∧C2vision) ⇒ [!(≥ 1)][!(nobody knows)]k−1 (da ⇒ 2a (¬da ⇒ dirties know)),
from which you can derive
(exactly k+1∧C2vision) ⇒ [!(≥ 1)][!(nobody knows)]k−1 (da ⇒ 2a [!(nobody knows)]da ).
From this (using the explicitation of “dirties know ”), prove the
desired inductive conclusion (case k + 1):
(exactly k+1 ∧ C2vision) ⇒ [!(≥ 1)][!(nobody knows)]k dirties know .
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