Certification Problem

Input (COPS 421)

We consider the TRS containing the following rules:

f(x,h(x)) f(h(x),h(x)) (1)
f(x,k(y,z)) f(h(y),h(y)) (2)
h(x) k(x,x) (3)
k(a,a) h(b) (4)
a b (5)

The underlying signature is as follows:

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

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:

f(x,k(y,z)) f(h(y),h(y)) (2)
h(x) k(x,x) (3)
a 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.