Date: Sat, 28 Feb 2004 13:43:27 -0500 (EST)
From: Michael Bloodgood <bloodgoo@mail.eecis.udel.edu>
To: Dave Saunders <saunders@mail.eecis.udel.edu>
Subject: 603 hw1 model solutions



I have attached a file (hw1.txt) containing model solutions for the
problems that had relatively frequent errors.

Mike

This document contains model solutions for the problems from hw1 that were
frequently done incorrectly.  

2.4

~f 
= ~f v f  (or-simp)
= t (excluded middle)

4.2.4

(N i: 0 <= i < n : b[i]=x) = (N j: 0 <= j < m: c[j]=x)

4.2.5

(A x (N i: 0 <= i < n: b[i]=x) = (N i: 0 <= i < n: c[i]=x))

4.4.3

E^i_j = E is valid

E^n_k = (A i: 1 <= i < k: (E j: b[j] = i)) is valid

For the third one, I accepted two answers.  
1: you could say it was invalid on account of the i being captured. 

2: you could say it is valid as long as the original i was changed to an unused
var first to avoid conflict and then the substitution was performed.  
e.g. a k could be substituted for the original i yielding

E^n_(n*i) = (A k: 1 <= k < n*i: (E j: b[j] = k)) is valid.  


Extra Credit (2.10):

Prove: if e is a tautology, then e=T can be proved to be an equivalence using
only the laws 1-12 and the rules of Substitution and Transitivity.  

(1) Suppose e taut.
(2) By 2.8, e = e1 where e1 is in CNF.
(3) e=e1 is arrived at from e by using only laws 1-12, Sub, and Trans.  This is
justified by the proof of 2.8.  
(4) e and e1 have the same truth value in every state (by (2), Def. of '='
(meaning))
(5) e1 and e have the same identifiers  (by proof of 2.8)
(6) e,e1 are well-defined in the same set of states  (by (5))
(7) e1 taut (by (1),(4), and (6)
(8) Recall e1 is of the form e_0^...^e_n where each e_i has the form g_0v...vg_m
where each g_j is an id, ~id,t, or f and the ids in each e_i are distinct.  
Therefore, the truth values of e_0,...,e_n are independent of each other by the
fact that the ids in each e_i are distinct.  By (7) and truth tables for ^, we
know that each of e_0,...,e_n must always evaluate to t when well-defined.  So
each e_i must be of the form g_0v...vg_m where at least one of the following two
cases holds:
Case A: at least one of the g_i=t
Case B: there exists g_i and g_j (i != j) s.t (such that) g_i = id_x and g_j=
~id_x.  

Now it is clear that from e1 by a suitable number of applications of Excluded
Middle (for the B cases) and Or-Simplification (for the A cases) we can prove e1
= T ^ ... ^ T.  Then we can apply And-Simplification next to arrive at e1=T 
(with an implicit uses of Sub and Trans as many times as needed throughout this
proof of e1=t). 

(9) In conclusion, by (8),(2),and (3), we have shown that e=T can be proved to
be an equivalence using only the laws 1-12, Sub, and Trans.  

(10) BY (1) and (9), the proof is now complete.