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 2023)

1 Redundant Rules Transformation

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

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)

All redundant rules that were added or removed can be simulated in 4 steps .

1.1 Critical Pair Closing System

Confluence is proven using the following terminating critical-pair-closing-system R:

k2(b) d (4)
k1(a) c (3)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c] = 0
[k2(x1)] = 2 · x1 + 0
[b] = 0
[d] = 0
[a] = 0
[k1(x1)] = 4 · x1 + 1
all of the following rules can be deleted.
k1(a) c (3)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[k2(x1)] = 1 · x1 + 1
[b] = 4
[d] = 0
all of the following rules can be deleted.
k2(b) d (4)

1.1.1.1.1 R is empty

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