Certification Problem

Input (COPS 87)

We consider the TRS containing the following rules:

f(x) → g(f(x)) (1)
h(x) → p(h(x)) (2)
f(x) → h(f(x)) (3)
g(x) → p(p(h(x))) (4)

The underlying signature is as follows:

{f/1, g/1, h/1, p/1}

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:

h(x) → p(h(x)) (2)
f(x) → h(f(x)) (3)
g(x) → p(p(h(x))) (4)

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:

There are no rules.

1.1.1 R is empty

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