OK, let me show it in another way.
Suppose I have one higher level specification:
VARIABLE a \* a is an integer
Next == \/ otherSubActions
\/ a' = a + 1
And now I want to design a implementation which refines the above specification, but in the implementation it batches the a' = a + 1 steps:
VARIABLE a \* a is an integer
Next == \/ otherSubActions
\/ a' = a + 10
In the implementation the variable "a" always increases 10 at a time, which means batching 10 steps of the specification's subAction2.
I think this is actually a kind of refinement, but I can not use stuttering variables to prove it, because : after each step of Specification, the variable "a" will change, while stuttering steps don't change any external state.
On Friday, March 23, 2018 at 9:45:06 PM UTC+8, Stephan Merz wrote:
Hello,
I have a hard time understanding the question. For one thing, the parameter a of subaction1 appears to be a variable (since you use a') but in subaction2 the quantified variable a denotes a value, and therefore a' = a.
Also, when you write a specification that involves sets of values, you typically use functions and have actions with conjuncts of the form
x' = [x EXCEPT ![a] = ...]
Then, if you try to combine several of those you obtain
\A a \in A : x' = [x EXCEPT ![a] = ...]
which is contradictory when A has more than one element.
Perhaps you could share a more complete specification to clarify what you are trying to achieve.
Regards,
Stephan
I am trying to prove one algorithm refines another algorithm. Let's call them Implementation and Specification.
For example, the Specification has one subaction that :
subaction1(a) == /\ F1(a)
/\ F2(a')
While in the Implementation, it will batch several of this step in one subaction, which is like:
subaction2(A) ==
\A a \in A :
/\ F1(a) /\ F2(a')
(A is a set consisting of element a, and each element doesn't interfere with other elements in A)
Then how can I represent it with the format of refinement mapping, for example using auxiliary variables? Or is it not a refinement mapping cause they doesn't have same behavior?
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