Certification Problem

Input (COPS 50)

We consider the TRS containing the following rules:

F(G(x,A,B)) x (1)
G(F(H(C,D)),x,y) H(K1(x),K2(y)) (2)
K1(A) C (3)
K2(B) D (4)

The underlying signature is as follows:

{F/1, G/3, A/0, B/0, H/2, C/0, D/0, K1/1, K2/1}

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:

K2(B) D (4)
K1(A) C (3)
G(F(H(C,D)),x,y) H(K1(x),K2(y)) (2)
F(G(x,A,B)) 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:

K1(A) C (3)
K2(B) D (4)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[C] = 0
[K2(x1)] = 4 · x1 + 1
[B] = 0
[D] = 0
[A] = 0
[K1(x1)] = 2 · x1 + 0
all of the following rules can be deleted.
K2(B) D (4)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[C] = 0
[A] = 4
[K1(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
K1(A) C (3)

1.1.1.1.1 R is empty

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