Invariant Generation Constraint-based approach - system

Question

I have been read the paper "Constraint-Based Linear-Relations Analysis" from "Sriram Sankaranarayanan, Henny B. Sipma, and Zohar Mann" to check fixpoint equations arising from abstract interpretation by applying Farkas Lemma in a Given template inequality with unknown coefficients, which computes constraints on the values of the coefficients, such that substituting any solution back into the template yields a valid invariant relationship.
I have Followed example 1 (on that paper)
Let V = {x, y} and L = { 0 }. Consider the transition system shown below. Each transition models a concurrent process, that updates the variables x, y automatically.
Θ = (x = 0 ∧ y = 0)
T = {τ1 , τ2 }
τ1 = <l0 , l0 , [x' = x + 2y ∧ y' = 1 − y]>
τ2 = <l0, l0 , [x' = x + 1 ∧ y' = y + 2]>
I have encoded Initiation, by using Farkas Lemma (example 2 on that paper) as well consecution (through transitions τ1 and τ2).
The authors say:
We fix a linear transition system Π with variables {x1 , . . . , xn }, collectively referred to as x. The system is assumed to have a single location to simplify the presentation. The template assertion at location , is α(c) = c1 x1 + · · · + cn xn + d ≥ 0. The coefficient variables {c1 , . . . , cn , d} are collectively referred to as c. The system's transitions are {τ1 , . . . , τm }, where τi : , , ρi. The initial condition is denoted by Θ. The system in Example 1 will be used as a running example to illustrate the presented ideas.
I have reached the overall constraint obtained by the conjunction of the constraints obtained from initiation and consecution for each transition (example 4 on that paper).
At that point I guess it is possible to solve the constraint by encoding all of that in a solve like Z3.
In fact I did that by encoding linear arithmetic directly into Z3:
(define-sort MyType () Int)
(declare-const myzero MyType)
(declare-const mi1 MyType)
(declare-const mi2 MyType)
(declare-const c1 MyType)
(declare-const c2 MyType)
(declare-const d MyType)
(assert (= myzero 0))
;initiation
(assert (>= d 0) )
;transition 1
(assert (and
(= (- (* mi1 c1) c1) 0)
(= (- (+ (* mi1 c2) c2) (* 2 c1) ) 0)
(<= (- (- (* mi1 d) d) c2) 0)
(>= mi1 0)
))
;transition 2
(assert (and
(= (- (* mi2 c1) c1) 0)
(= (- (* mi2 c2) c2) 0)
(<= (- (- (- (* mi2 d) d) c1) (* 2 c2) ) 0)
(>= mi2 0)
))
(check-sat)
(get-model)
I guess I am not doing well once I have not figured out any inductive invariant at location l0 through c1, c2, ..., cn, d as an individual (o range) values.
Z3 has answered me zero for all coefficients:
sat
(model
(define-fun mi2 () Int 0)
(define-fun c2 () Int 0)
(define-fun mi1 () Int 0)
(define-fun c1 () Int 0)
(define-fun d () Int 4)
(define-fun myzero () Int 0)
)
I have tried to found examples related with, but until now no luck to get it.

Answer 1

If I understand what you are doing correctly in your Z3 encoding, you are indeed
getting correct but trivial invariants like 0 <= 4.
To get interesting invariants, I suggest adding constraints like c1 <> 0 to see if the solver gives you something interesting back.
Our work was done long before Z3 existed: we used the solver REDLOG as part of REDUCE, which is still available. Welcome to email me with your queries.
Best,
Sriram