Certification Problem

Input (COPS 48)

We consider the TRS containing the following rules:

H(F(x,y)) → F(H(R(x)),y) (1)
F(x,K(y,z)) → G(P(y),Q(z,x)) (2)
H(Q(x,y)) → Q(x,H(R(y))) (3)
Q(x,H(R(y))) → H(Q(x,y)) (4)
H(G(x,y)) → G(x,H(y)) (5)

The underlying signature is as follows:

{H/1, F/2, R/1, K/2, G/2, P/1, Q/2}

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:

H(G(x,y)) → G(x,H(y)) (5)
Q(x,H(R(y))) → H(Q(x,y)) (4)
H(Q(x,y)) → Q(x,H(R(y))) (3)
F(x,K(y,z)) → G(P(y),Q(z,x)) (2)
H(F(x,y)) → F(H(R(x)),y) (1)
Q(x,H(R(y))) → Q(x,H(R(y))) (6)
H(Q(x,y)) → H(Q(x,y)) (7)

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

1.1 Strongly closed

Confluence is proven since the TRS is strongly closed. The joins can be performed using 7 step(s).