Certification Problem

Input (COPS 34)

We consider the TRS containing the following rules:

f(a,b) c (1)
a a' (2)
b b' (3)
c f(a',b) (4)
c f(a,b') (5)
c f(a,b) (6)

The underlying signature is as follows:

{f/2, a/0, b/0, c/0, a'/0, b'/0}

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 a' (2)
b b' (3)
c f(a,b) (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.