Certification Problem

Input (COPS 763)

We consider the TRS containing the following rules:

f(f(x)) → f(g(f(x),f(x))) (1)

The underlying signature is as follows:

{f/1, g/2}

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(f(x)) → f(g(f(x),f(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:

f(f(x)) → f(g(f(x),f(x))) (1)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[g(x1, x2)] =
2 0
0 0
· x1 +
1 0
0 0
· x2 +
0 0
0 0
[f(x1)] =
1 2
1 2
· x1 +
0 0
1 0
all of the following rules can be deleted.
f(f(x)) → f(g(f(x),f(x))) (1)

1.1.1.1 R is empty

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