Certification Problem

Input (COPS 951)

We consider the TRS containing the following rules:

0(1(2(3(4(x))))) → 0(2(1(3(4(x))))) (1)
0(5(1(2(4(3(x)))))) → 0(5(2(1(4(3(x)))))) (2)
0(5(2(4(1(3(x)))))) → 0(1(5(2(4(3(x)))))) (3)
0(5(3(1(2(4(x)))))) → 0(1(5(3(2(4(x)))))) (4)
0(5(4(1(3(2(x)))))) → 0(5(4(3(1(2(x)))))) (5)

The underlying signature is as follows:

{0/1, 1/1, 2/1, 3/1, 4/1, 5/1}

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:

0(5(4(1(3(2(x)))))) → 0(5(4(3(1(2(x)))))) (5)
0(5(3(1(2(4(x)))))) → 0(1(5(3(2(4(x)))))) (4)
0(5(2(4(1(3(x)))))) → 0(1(5(2(4(3(x)))))) (3)
0(5(1(2(4(3(x)))))) → 0(5(2(1(4(3(x)))))) (2)
0(1(2(3(4(x))))) → 0(2(1(3(4(x))))) (1)

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