Certification Problem

Input (COPS 85)

We consider the TRS containing the following rules:

f(g(x)) h(g(x),g(x)) (1)
f(s(x)) h(s(x),s(x)) (2)
g(x) s(x) (3)

The underlying signature is as follows:

{f/1, g/1, h/2, s/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:

g(x) s(x) (3)
f(s(x)) h(s(x),s(x)) (2)
f(g(x)) h(g(x),g(x)) (1)
f(g(x)) h(g(x),s(x)) (4)
f(g(x)) h(s(x),g(x)) (5)

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:

f(s(x)) h(s(x),s(x)) (2)
g(x) s(x) (3)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[f(x1)] = 5 · x1 + 4
[s(x1)] = 2 · x1 + 2
[g(x1)] = 4 · x1 + 4
[h(x1, x2)] = 4 · x1 + 1 · x2 + 4
all of the following rules can be deleted.
g(x) s(x) (3)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[f(x1)] = 2 · x1 + 5
[s(x1)] = 1 · x1 + 0
[h(x1, x2)] = 1 · x1 + 1 · x2 + 0
all of the following rules can be deleted.
f(s(x)) h(s(x),s(x)) (2)

1.1.1.1.1 R is empty

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