Certification Problem

Input (COPS 96)

We consider the TRS containing the following rules:

F(H(x),y) → F(H(x),I(I(y))) (1)
F(x,G(y)) → F(I(x),G(y)) (2)
I(x) → x (3)

The underlying signature is as follows:

{F/2, H/1, I/1, G/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:

I(x) → x (3)

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.