Certification Problem

Input (COPS 286)

The rewrite relation of the following conditional TRS is considered.

f(x,x)	→	a
g(x)	→	a	\| g(x) ≈ b
a	→	b
b	→	a

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

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
g(x) → a | g(x) ≈ b
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 CTRS without Conditions

Switching from confluence of a CTRS without conditions to confluence of the corresponding TRS.

1.2.1 Redundant Rules Transformation

To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:

f(x,x)	→	a	(2)
b	→	a	(3)

All redundant rules that were added or removed can be simulated in 4 steps .

1.2.1.1 Locally confluent and terminating

Confluence is proven by showing local confluence and termination.

1.2.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals

[a]	=	0
[f(x₁, x₂)]	=	1 · x₁ + 6 · x₂ + 0
[b]	=	4

all of the following rules can be deleted.

→

(3)

1.2.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals

[a]	=	0
[f(x₁, x₂)]	=	1 · x₁ + 4 · x₂ + 1

all of the following rules can be deleted.

f(x,x)

→

(2)

1.2.1.1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.

1.2.1.1.2 Local Confluence Proof

All critical pairs are joinable which can be seen by computing normal forms of all critical pairs.