Certification Problem

Input (COPS 111)

We consider the TRS containing the following rules:

a b (1)
a c (2)
a e (3)
b d (4)
c a (5)
d a (6)
d e (7)
g(x) h(a) (8)
h(x) e (9)

The underlying signature is as follows:

{a/0, b/0, c/0, e/0, d/0, g/1, h/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:

a e (3)
b d (4)
c a (5)
d e (7)
g(x) h(a) (8)
h(x) e (9)

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.