Certification Problem

Input (COPS 693)

We consider the TRS containing the following rules:

c → b (1)
a → a (2)
b → b (3)
f(f(a)) → c (4)

The underlying signature is as follows:

{c/0, b/0, a/0, f/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:

f(f(a)) → c (4)
c → b (1)

All redundant rules that were added or removed can be simulated in 1 steps .

1.1 Redundant Rules Transformation

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

f(f(a)) → c (4)
c → b (1)
f(f(a)) → b (5)

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

1.1.1 Critical Pair Closing System

Confluence is proven using the following terminating critical-pair-closing-system R:

c → b (1)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b] = 0
[c] = 1
all of the following rules can be deleted.
c → b (1)

1.1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.