Certification Problem

Input (COPS 419)

We consider the TRS containing the following rules:

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

The underlying signature is as follows:

{f/2, h/1, g/1}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2020)

1 Redundant Rules Transformation

To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:

f(x,y) f(y,x) (3)
h(x) g(x) (5)
g(x) x (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.