Certification Problem

Input (COPS 88)

We consider the TRS containing the following rules:

p(x) q(x) (1)
p(x) r(x) (2)
q(x) s(p(x)) (3)
r(x) s(p(x)) (4)
s(x) f(p(x)) (5)

The underlying signature is as follows:

{p/1, q/1, r/1, s/1, f/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:

p(x) r(x) (2)
q(x) s(p(x)) (3)
r(x) s(p(x)) (4)
s(x) f(p(x)) (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.