Model Solutions for hw2.

1. Deduce: Law of Mon. using (7.3),(7.4) and basic laws in Ch. 2.  

Suppose the hyp. q->r.  

w(q)->w(r)
= w(q^r)->w(r) by hyp. and metathm
= (w(q)^w(r))->w(r)  by Dist. of Conj.
= ~(w(q)^w(r)) v w(r) by Implication
= ~w(q) v ~w(r) v w(r) by DeMorgan
= ~w(q) v T by Ex. Mid.
= T by Or-Simp.

Deduce: Dist. of Disj. using (7.3),(7.4), and basic laws.  
w(q) v w(r) -> w(q v r)
= ~(w(q) v w(r)) v w(q v r) by Imp.
= (~w(q) ^ ~w(r)) v w(q v r) by DeMorgan
= (~w(q) v w (q v r)) ^ (~w(r) v w(q v r)) by Dist.
= (~w(q ^ (q v r)) v w(q v r)) ^ (~w(r) v w(q v r)) by And-Simp.
= (~(w(q) ^ w(q v r)) v w(q v r)) ^ (~w(r) v w(q v r)) by Dist. of Conj.
= (~w(q) v ~w(q v r) v w(q v r)) ^  (~w(r) v w(q v r)) by DeMorgan
= (~w(q) v T) ^ (~w(r) v w (q v r)) Ex. Middle
= T ^ (~w(r) v w(q v r)) by or-simp.
= ~w(r) v w(q v r) by and-simp.
= ~w(r^(q v r)) v w(q v r) by and-simp
= ~(w(r) ^ w(q v r)) v w(q v r) by dist. of conj.
= ~w(r) v ~w(q v r) v w(q v r) by DeMorgan
= ~w(r) v T  by Ex. Middle
= T or-simp.

7.5 There are many possibilities.  One example is let S=abort.  Then 
wp(abort,R) v wp(abort, ~R)
= f v f by def of abort
= f by or-simp

8.2

(7.3)
wp(abort, f) = f by def of abort

(7.4)
wp(abort,q) ^ wp(abort,r)
= f ^ f by def of abort
= f by and-simp
= wp(abort,q ^ r) by def of abort

(7.7)
wp(abort,q) v wp(abort,r)
=f v f by def of abort
=f  by or-simp
=wp(abort,q v r) by def of abort

8.5

TST (To Show That) : wp("s1;s2",q) v wp("s1;s2",r) = wp("s1;s2",q v r) given
that s1 and s2 satisfy 7.7.

wp("s1;s2",q) v wp("s1;s2",r)
= wp(s1,wp(s2,q)) v wp(s1,wp(s2,r)) by def. 8.3
= wp(s1,wp(s2,q) v wp(s2,r)) by 7.7 and since s1 satisfies 7.7
= wp(s1,wp(s2,q v r)) by 7.7 and since s2 satisfies 7.7
= wp("s1;s2", q v r) by def. 8.3

9.2.1
Most of them got this.  

9.2.2
Consider R = y>0 ^ x >=0, e1 = -y, e2=x, and state s = {x=1,y=0}
wp(x,y:=-y,x,y>0^x>=0)
= x>0 ^ -y >= 0  by def of multiple asgnmnt.  In s, this is T.  

wp(x:=-y,y:=x,y>0 ^ x>=0)
= wp(x:=-y,wp(y:=x,y>0^x>=0)) by def of sequential
= wp(x:=-y,x>0 ^ x>=0) by def of asnmnt and text sub
= -y>0 ^ -y >=0 by def of asnmnt and text sub.  In s, this is F.

wp(y:=x,x:=-y,y>0^x>=0)
=wp(y:=x,wp(x:=-y,y>0^x>=0)) by def of seq
=wp(y:=x,y>0^-y>=0) by def of asnmnt and text sub.
= x>0 ^-x>=0 by def of asnmnt and text sub.  In s, this is F.  

10.2

Show IF satisfies (7.3) provided the subcommands do.  

wp(IF,F)
= BB ^ (A i: 1 <=i<=n:Bi=>wp(Si,F))   by def of IF and in the future of this
proof I will refer to this expression as *

There are 2 cases to consider.  

Case 1: Every Bi is F.  Then we have *
= F ^ (A i: 1 <= i <= n: Bi=>wp(Si,R)) by or-simp
= F by and-simp.

Case 2: ow. (not every Bi is F)

Then we have *
= T ^ (A i: 1<=i<=n:Bi=>wp(Si,F)) or-simp.
= T ^ (A i: 1<=i<=n:Bi=>F) because we're given that the subcommands satisfy
(7.3)

= T ^ ((~B1 v F) ^ (~B2 v F) ^ ... ^ (~Bn v F)) expand quantifier and by law of
implication

= T ^ (~B1 ^ ~B2 ^ ... ^ ~Bn) by or-simp.
= T ^ ~T ^ (...) by Comm., Assoc., and Description of this Case
= F by Contradiction, And-Simp.

Show IF satisfies (7.4) provided the subcommands do. 

wp(IF, q) ^ wp(IF, r)
=bb ^ (b1=>wp(s1,q)^...^bn=>wp(sn,q)) ^ bb ^ (b1=>wp(s1,r)^...^bn=>wp(sn,r)) by
def of IF

= bb ^ (b1=>wp(s1,q)^b1=>wp(s1,r)^...^bn=>wp(sn,q)^bn=>wp(sn,r)) by and-simp.,
comm., and assoc.

= bb ^ (b1=>wp(s1,q^r)^...^bn=>wp(sn,q^r)) by Lemma (see below) and given that
subcommands satisfy (7.4)

= wp(IF,q^r) by def of IF

Lemma: b1=>a ^ b1=>b
= b1=>a^b

proof of Lemma:
b1=>a ^ b1=>b
=(~b1 v a) ^ (~b1 v b) by law of implication
= ~b1 v (a ^ b) by Distributive law
= b1=>a^b  by Implication

10.4 

wp(s4,a>0^b>0)
=(a>b v b>a) ^ (a>b)=>wp(a:=a-b,a>0^b>0) ^ (b>a)=>wp(b:=b-a,a>0^b>0) by def of
IF

= (a>b v b>a) ^ (a>b)=>(a-b>0 ^ b>0) ^ (b>a)=>(a>0 ^ b-a>0) by def of asnmnt and
text sub.

= (a!=b) ^ (a>0 ^ b>0) by Math Simplification

10.7
Using Thm 10.5, prove that the following proof outline is t.  
q: {x>0 ^ z+y*x=a*b}
s1:if odd(x)->z,x:=z+y,x-1
   || even(x)->skip fi;
s2: y,x:=2*y,x/2
r: {x>=0 ^ z+y*x=a*b}

wp("s1;s2",r)
= wp(s1,wp(s2,r)) by def of seq.
= wp(s1,wp(y,x:=2*y,x/2,x>=0 ^ z+y*x=a*b) by Sub.
= wp(s1,(x/2>=0 ^ z+2*y*x/2=a*b)) by def of multiple asnmnt and text sub.
= wp(s1,(x>=0 ^ z+y*x=a*b)) simplifying x/2 <=> x>=0 and cancel the 2's
Now, by Thm 10.5 it will suffice TST:
(1) q=>odd(x) v even(x) = t
(2) q ^ odd(x) =>wp(z,x:=z+y,x-1,x>=0^z+y*x=a*b)= t
(3) q ^ even(x) =>wp(skip,x>=0^z+y*x=a*b)= t

TST (1):
q=>odd(x) v even(x)
=~q v (odd(x) v even(x)) by Implication
= ~q v t by Ex. Middle and odd(x) is the negation of even(x)
= t by or-simp.

TST (2):
q ^ odd(x) => wp(z,x:=z+y,x-1,x>=0 ^ z+y*x=a*b)
= q ^ odd(x) => (x-1>=0 ^ z+y+y*(x-1)=a*b) by def of mult. asnmnt and text sub.
= q ^ odd(x) => (x>=0 ^ z+y*x=a*b) by simplification: x-1>=0 <=> x>0 and cancel
the y's
= ~(q ^ odd(x)) v q by substituting q and implication
= ~q v ~odd(x) v q DeMorgan
= t by Ex. Middle and or-simp.

TST (3):
q ^ even(x) => wp(skip,x>=0 ^ z+y*x=a*b)
= q ^ even(x) => (x>=0 ^ z+y*x=a*b) by def of skip
= x>0 ^ z+y*x=a*b ^ even(x) => (x>=0 ^ z+y*x=a*b) substituting for q
= ~x>0 v ~(z+y*x=a*b) v ~even(x) v ((x>0 v x=0) ^ z+y*x=a*b) by Implication,
DeMorgan, x>=0 <=> x>0 v x=0
= ~(x>0) v ~(z+y*x=a*b) v ~even(x) v (x>0 ^ z+y*x=a*b) v (x=0 ^ z+y*x=a*b) by
Dist. Law
= ~(x>0 ^ z+y*x=a*b) v ~even(x) v (x>0 ^ z+y*x=a*b) v (x=0 ^ z+y*x=a*b) by
Assoc, DeMorgan
= t by Comm, Excl. Middle, or-simp

11.1

wp(do od, R)
= (E k: 0 <= k: H_k(R)) by Def. of DO
H_0(R) = ~bb ^ R = R since by definition, the ~emptydisjunction = t and by
And-simp.
Proceeding by Induction, H_k(R) = R v wp(if fi, H_(k-1)(R))
= R v wp(if fi,R) by Induction
= R v f, by 10.1 (done in class)
= R by or-simp.
So (E k: 0 <= k: H_k(R)) 
= (E k: 0 <= k: R)
= R.  

It resembles a skip command.