Certification Problem

Input (COPS 1651)

We consider the TRS containing the following rules:

f(g(h(x))) → g(f(h(g(x)))) (1)
f(x) → x (2)
g(x) → x (3)
h(x) → x (4)

The underlying signature is as follows:

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

h(x) → x (4)
g(x) → x (3)
f(x) → x (2)
f(g(h(x))) → g(f(h(g(x)))) (1)
f(g(h(x))) → g(f(g(x))) (5)
f(g(h(x))) → g(f(h(x))) (6)
f(g(h(x))) → f(h(g(x))) (7)
f(g(h(x))) → g(h(g(x))) (8)

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:

g(x) → x (3)
h(x) → x (4)
f(x) → x (2)

1.1.1 Rule Removal

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

1.1.1.1 R is empty

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