Assignment #2: COS 441

Due by midnight on Wednesday Oct. 1.  Either hand in in class or
deliver to Rob's office.  You mail also email a completed assignment
to him at rdockins@princeton.edu.  Please put "COS 441 A2" in the
subject line of any such email.


There are 15 points to gain on this assignment.

Question 1:  Remember our natural numbers and our lists:

n ::= Z | S Z

l ::= nil | cons (n,l)

Here is a definition of the less-than-or-equal-to judgement for
natural numbers:

Judgement Form:  |- leq n1 n2

Rules:

------------ (Z-LEQ)
|- leq Z n2 


|- leq n1 n2
--------------------- (S-LEQ)
|- leq (S n1) (S n2) 

(a) [1 point]  

Use the leq judgement to define a new judgement with the form

   |- ascend l 

that is valid whenever the elements of l are in ascending order
(duplicates are allowed). For example, these judgements are valid:

|- ascend cons (Z, cons (Z, cons (S S S Z, cons (S S S S S S Z, nil))))

|- ascend nil

|- ascend cons (S S Z, nil)

This judgement is not valid:

|- ascend cons (Z, cons (S S Z, cons ( S Z, nil)))

(b) [2 points]  Define a judgement with the form:

|- reverse l1 l2

where the elements in l2 appear in the reverse order as in l1. eg:

|- reverse (cons (Z, cons (S Z, nil)))  (cons (S Z, cons (Z, nil)))

(c) [4 points]
Consider the judgement |- dup l1 l2 and its rules:

---------------(nil-dup)
|- dup nil nil


|- dup l1 l2
----------------------------------------- (cons-dup)
|- dup cons(n,l1)    cons(n, cons(n, l2)) 

Prove this theorem:
If |- ascend l1 and |- dup l1 l2 then |- ascend l2.

(d) [1 point]  Define a judgement |- incr l1 l2 where l2 is the same as
l1 except you have added one to every element (ie, gone from n to (S n) for
every element n of the list).  For example, this judgement is valid:

|- incr (cons (Z,cons (S S S Z, nil))) (cons (S Z, cons (S S S S Z, nil)))

(e) [4 points]
Prove this theorem:  
If |- ascend l1 and |- incr l1 l2 then |- ascend l2.


Question 2:

Here is the definition of a very simple logic with 
true, false, inequalities, conjunction and disjunction.  
The formulas F are the following:

F ::= true | false | n1 <= n2 | F1 /\ F2 | F1 \/ F2

The truth of a logical formula is determined by a judgement with the form:

|- F 

Here are the rules: 

-------- (T)
|- true

|- leq n1 n2
-------------- (LEQ)
|- n1 <= n2

|- F1     |- F2                
-----------------------  (/\)         
|- F1 /\ F2                           

|- F1                        |- F2
------------------ (\/1)     ----------- (\/2)        
|- F1 \/ F2                  |- F1 \/ F2
    

(a) [1 Point]  Write down a complete derivation for the judgement:

|- (false \/ ((Z <= S Z) \/ (S Z <= Z))) /\ true

(b)  [1 Point] Prove that: 
if   |- (F1 /\ (F2 /\ F3)) is a valid derivation (for any F1, F2, F3)
then |- ((F2 /\ F3) /\ F1)) is a valid derivation.
Hint: Don't prove it by induction.  Explain what the derivation of
|- (F1 /\ (F2 /\ F3)) must look like and why.  Then use that information
to construct a derivation for |- ((F2 /\ F3) /\ F1)).

(c) [1 Point] Prove that: 
if   |- (F1 \/ (F2 \/ F3)) is a valid derivation (for any F1, F2, F3)
then |- ((F2 \/ F3) \/ F1)) is a valid derivation.
Hint:  Again, don't use induction.  Consider the possible forms that
the derivation of |- (F1 \/ (F2 \/ F3)) could have and go from there.