Certification Problem

Input (COPS 92)

We consider the TRS containing the following rules:

W(B(x))	→	W(x)	(1)
B(I(x))	→	J(x)	(2)
W(I(x))	→	W(J(x))	(3)

The underlying signature is as follows:

{W/1, B/1, I/1, J/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:

W(I(x))	→	W(J(x))	(3)
B(I(x))	→	J(x)	(2)
W(B(x))	→	W(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:

W(I(x))

→

W(J(x))

(3)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals

[W(x₁)]	=	4 · x₁ + 2
[J(x₁)]	=	1 · x₁ + 1
[I(x₁)]	=	1 · x₁ + 2

all of the following rules can be deleted.

W(I(x))

→

W(J(x))

(3)

1.1.1.1 R is empty

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