Certification Problem

Input (COPS 423)

We consider the TRS containing the following rules:

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

The underlying signature is as follows:

{g/2, f/1, h/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) g(x,f(x)) (2)
h(x) g(f(x),x) (3)
g(x,y) h(g(f(x),f(y))) (4)

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.