Certification Problem

Input (COPS 71)

We consider the TRS containing the following rules:

F(x,y) → c(y) (1)
G(x) → x (2)
f(x) → g(x) (3)
g(x) → c(x) (4)

The underlying signature is as follows:

{F/2, c/1, G/1, f/1, g/1}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2023)

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 : 2 ⨯ 3 → 0
c : 3 → 0
G : 1 → 1
f : 3 → 0
g : 3 → 0
The subsystems are

(1.1)

F(x,y) → c(y) (1)
f(x) → g(x) (3)
g(x) → c(x) (4)

(1.2)

G(x) → x (2)

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 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.