Certification Problem

Input (COPS 65)

We consider the TRS containing the following rules:

h(f,a,a) h(g,a,a) (1)
h(g,a,a) h(f,a,a) (2)
a a' (3)
h(x,a',y) h(x,y,y) (4)
g f (5)
f g (6)

The underlying signature is as follows:

{h/3, f/0, a/0, g/0, a'/0}

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:

a a' (3)
h(x,a',y) h(x,y,y) (4)
f g (6)

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.