Certification Problem

Input (COPS 61)

We consider the TRS containing the following rules:

a → a' (1)
h(x,a',y) → h(x,y,y) (2)
h(x,y,a') → h(x,y,y) (3)
g → f (4)
h(f,a,a) → h(g,a,a) (5)
h(g,a,a) → h(f,a,a) (6)

The underlying signature is as follows:

{a/0, a'/0, h/3, g/0, f/0}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2021)

1 Redundant Rules Transformation

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

a → a' (1)
h(x,a',y) → h(x,y,y) (2)
h(x,y,a') → h(x,y,y) (3)
g → f (4)

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.