Certification Problem

Input (COPS 705)

We consider the TRS containing the following rules:

a	→	c	(1)
c	→	a	(2)
c	→	h(a,c)	(3)

The underlying signature is as follows:

{a/0, c/0, h/2}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2022)

1 Redundant Rules Transformation

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

c	→	h(a,c)	(3)
c	→	a	(2)
a	→	c	(1)
a	→	h(a,c)	(4)

All redundant rules that were added or removed can be simulated in 2 steps .

1.1 Decreasing Diagrams

1.1.2 Rule Labeling

Confluence is proven, because all critical peaks can be joined decreasingly using the following rule labeling function (rules that are not shown have label 0).

c → h(a,c) (3)
↦ 1
c → a (2)
↦ 0
a → c (1)
↦ 2
a → h(a,c) (4)
↦ 0

The critical pairs can be joined as follows. Here, ↔ is always chosen as an appropriate rewrite relation which is automatically inferred by the certifier.

The critical peak s = h(a,c)←→_ε a = t can be joined as follows.
s ↔ h(a,c) ↔ t
The critical peak s = a←→_ε h(a,c) = t can be joined as follows.
s ↔ t
The critical peak s = c←→_ε h(a,c) = t can be joined as follows.
s ↔ t
The critical peak s = h(a,c)←→_ε c = t can be joined as follows.
s ↔ h(a,c) ↔ t