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 2023)

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)
F(H(K1(A),K2(B))) → F(H(C,D)) (5)

All redundant rules that were added or removed can be simulated in 2 steps .

1.1 Development Closed

Confluence is proven since the TRS is development closed. The joins can be performed using 1 development step(s).