Certification Problem

Input (COPS 69)

We consider the TRS containing the following rules:

F(x,y) → c(A) (1)
G(x) → x (2)
h(x) → c(x) (3)

The underlying signature is as follows:

{F/2, c/1, A/0, G/1, h/1}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2022)

1 Persistent Decomposition (Many-Sorted)

Confluence is proven, because the maximal systems induced by the sorts in the following many-sorted sort attachment are confluent.
F : 3 ⨯ 4 → 0
c : 1 → 0
A : 1
G : 2 → 2
h : 1 → 0
The subsystems are

(1.1)

F(x,y) → c(A) (1)
h(x) → c(x) (3)

(1.2)

G(x) → x (2)

1.1 Redundant Rules Transformation

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

h(x) → c(x) (3)
F(x,y) → c(A) (1)

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

1.1.1 Critical Pair Closing System

Confluence is proven using the following terminating critical-pair-closing-system R:

There are no rules.

1.1.1.1 R is empty

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

1.2 Redundant Rules Transformation

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

G(x) → x (2)

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

1.2.1 Critical Pair Closing System

Confluence is proven using the following terminating critical-pair-closing-system R:

There are no rules.

1.2.1.1 R is empty

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