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.
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.