181 CS Formal Methods in Software Engineering
Example for Homework 6
$Date: 2001/04/17 22:13:31 $
Here's an example of the kind of solution we're asking for in homework
6 for the loop problems. It's based on Cohen's section 10.2 (thanks
to Kevin Lillis for suggesting this).
(Step 0) We calculate P /\ !B ==> R as follows. Recall that P is P0 /\ P1.
Assume !(n != #b), i.e., n == #b.
P0 /\ P1
<==> <def of P0 and P1>
x == (\min i : 0 <= i /\ i < n : b.i) /\ 0 <= n /\ n <= #b
<==> <predicate calculus>
x == (\min i : 0 <= i /\ i < n : b.i) /\ 0 <= n /\ n <= #b
<==> <assumption that n == #b>
x == (\min i : 0 <= i /\ i < #b : b.i) /\ 0 <= #b /\ #b <= #b
==> <X /\ Y ==> X>
x == (\min i : 0 <= i /\ i < #b : b.i)
<==> <def of R>
R
(Step 1) We calculate S0 so that {Q} S0 {P} as folows. Assume (Q: #b >= 0).
wp.(n,x := NV, XV).(P0 /\ P1)
<==> <definition of assignment>
(P0 /\ P1)(n,x := NV, XV)
<==> <definition of P0 and P1>
XV == (\min i : 0 <= i /\ i < NV : b.i) /\ 0 <= NV /\ NV <= #b
<==> <choose NV := 0, since counting up, and to truthify 0 <= NV, etc.>
XV == (\min i : 0 <= i /\ i < 0 : b.i) /\ 0 <= 0 /\ 0 <= #b
<==> <arithmetic and definition of #b>
XV == (\min i : 0 <= i /\ i < 0 : b.i) /\ true
<==> <empty range rule for \min>
XV +inf
<==> == <choose XV := +inf>
true
so S0 is n,x := 0, +inf
(Step 2) We calculate P /\ t <= 0 ==> !B as follows.
P0 /\ P1 /\ t <= 0
<==> <definition of P1, and t>
P0 /\ 0 <= n /\ n <= #b /\ (#b - n) <= 0
==> <X /\ Y ==> Y>
0 <= n /\ n <= #b /\ (#b - n) <= 0
<==> <arithmetic>
0 <= n /\ n <= #b /\ #b <= n
<==> <antisymmetry>
0 <= n /\ n == #b
==> <X /\ Y ==> Y>
n == #b
<==> <idempotence of negation>
!(n != b)
<==> <definition of B>
!B
(Step 3a) The calculation of S so that {P /\ B} S {P} is on pages 156-157.
(Step 3b) We calculate {P /\ B /\ t == T} S {t < T} as follows.
Assume P0 /\ P1 /\ n != #b /\ (#b -n) == T.
wp.(n, x := n+1, x min b.n).(#b - n < T)
<==> <definition of assignment>...