Certification Problem

Input (COPS 66)

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)
f(h(k1(a),k2(b))) f(h(c,d)) (5)
f(h(c,k2(b))) f(h(c,d)) (6)
f(h(k1(a),d)) f(h(c,d)) (7)

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 2020)

1 Redundant Rules Transformation

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

f(h(k1(a),k2(b))) f(h(c,d)) (5)
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 1 steps .

1.1 Redundant Rules Transformation

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

f(h(k1(a),k2(b))) f(h(c,d)) (5)
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 .

1.1.1 Parallel Closed

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