Certification Problem

Input (COPS 938)

We consider the TRS containing the following rules:

a(b(c(x))) → c(c(c(b(b(b(a(a(a(x))))))))) (1)
c(b(x)) → a(a(a(x))) (2)
a(x) → x (3)
b(x) → x (4)
c(x) → x (5)

The underlying signature is as follows:

{a/1, b/1, c/1}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2023)

1 Redundant Rules Transformation

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

a(b(c(x))) → c(c(c(b(b(b(a(a(a(x))))))))) (1)
a(x) → x (3)
b(x) → x (4)
c(x) → x (5)

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

1.1 Critical Pair Closing System

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

c(x) → x (5)
b(x) → x (4)
a(x) → x (3)

1.1.1 Dependency Pair Transformation

The set of initial dependency pairs is empty, and hence the TRS is terminating.