Certification Problem

Input (COPS 72)

We consider the TRS containing the following rules:

F(c(x)) → G(x) (1)
G(x) → F(x) (2)
c(x) → x (3)

The underlying signature is as follows:

{F/1, c/1, G/1}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2021)

1 Redundant Rules Transformation

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

c(x) → x (3)
G(x) → F(x) (2)
F(c(x)) → G(x) (1)
F(c(x)) → F(x) (4)

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

1.1 Critical Pair Closing System

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

G(x) → F(x) (2)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[F(x1)] = 4 · x1 + 0
[G(x1)] = 4 · x1 + 1
all of the following rules can be deleted.
G(x) → F(x) (2)

1.1.1.1 R is empty

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