Certification Problem

Input (COPS 74)

We consider the TRS containing the following rules:

a c (1)
b c (2)
f(a,b) d (3)
f(x,c) f(c,c) (4)
f(c,x) f(c,c) (5)
d f(a,c) (6)
d f(c,b) (7)

The underlying signature is as follows:

{a/0, c/0, b/0, f/2, d/0}

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:

a c (1)
b c (2)
f(x,c) f(c,c) (4)
f(c,x) f(c,c) (5)
d f(c,b) (7)

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.