Certification Problem

Input (COPS 22)

We consider the TRS containing the following rules:

g(a) f(g(a)) (1)
g(b) c(a) (2)
a b (3)
f(x) h(x) (4)
h(x) c(b) (5)

The underlying signature is as follows:

{g/1, a/0, f/1, b/0, c/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:

g(b) c(a) (2)
a b (3)
f(x) h(x) (4)
h(x) c(b) (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:

There are no rules.

1.1.1 R is empty

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