Certification Problem

Input (COPS 271)

The rewrite relation of the following conditional TRS is considered.

p(q(x))	→	p(r(x))
q(h(x))	→	r(x)
r(x)	→	r(h(x))	\| s(x) ≈ 0
s(x)	→	1

Property / Task

Prove or disprove confluence.

Answer / Result

No.

Proof (by ConCon @ CoCo 2020)

1 Removal of Infeasible Rules

We may safely remove rules with infeasible conditions. They do not influence the rewrite relation in any way.

1.1 Rules with Infeasible Conditions

1.1.1 Rule with Infeasible Conditions
The rule
r(x) → r(h(x)) | s(x) ≈ 0
has infeasible conditions.
1.1.1.1 Infeasible Equation
The equation is infeasible.
1.1.1.1.1 Non-reachability
We show non-reachability w.r.t. the underlying TRS.
1.1.1.1.1.1 Non-reachability by TCAP
Non-reachability is shown by the TCAP approximation.

1.2 Non-Joinable Fork

The system is not confluent due to the following forking derivations.

p(q(h(z)))

→^*

p(r(z))

p(q(h(z)))

→^*

p(r(h(z)))

The two resulting terms cannot be joined for the following reason:

The terms are distinct normal forms.