Certification Problem

Input (COPS 803)

We consider the TRS containing the following rules:

g(a,y) y (1)
f(x,a) f(x,g(x,b)) (2)
g(h(x),y) g(x,h(y)) (3)

The underlying signature is as follows:

{g/2, a/0, f/2, b/0, h/1}

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:

g(h(x),y) g(x,h(y)) (3)
f(x,a) f(x,g(x,b)) (2)
g(a,y) y (1)

All redundant rules that were added or removed can be simulated in 2 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.