Certification Problem
Input (COPS 73)
We consider the TRS containing the following rules:
|
f(g(g(x))) |
→ |
a |
(1) |
|
f(g(h(x))) |
→ |
b |
(2) |
|
f(h(g(x))) |
→ |
b |
(3) |
|
f(h(h(x))) |
→ |
c |
(4) |
|
g(x) |
→ |
h(x) |
(5) |
| a |
→ |
b |
(6) |
| b |
→ |
c |
(7) |
The underlying signature is as follows:
{f/1, g/1, a/0, h/1, b/0, c/0}Property / Task
Prove or disprove confluence.Answer / Result
Yes.Proof (by csi @ CoCo 2021)
1 Redundant Rules Transformation
To prove that the TRS is (non-)confluent, we show (non-)confluence of the following
modified system:
| b |
→ |
c |
(7) |
| a |
→ |
b |
(6) |
|
g(x) |
→ |
h(x) |
(5) |
|
f(h(h(x))) |
→ |
c |
(4) |
|
f(h(g(x))) |
→ |
b |
(3) |
|
f(g(h(x))) |
→ |
b |
(2) |
|
f(g(g(x))) |
→ |
a |
(1) |
| a |
→ |
c |
(8) |
|
f(h(g(x))) |
→ |
c |
(9) |
|
f(g(h(x))) |
→ |
c |
(10) |
|
f(g(g(x))) |
→ |
b |
(11) |
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:
| b |
→ |
c |
(7) |
|
f(h(h(x))) |
→ |
c |
(4) |
|
f(h(g(x))) |
→ |
c |
(9) |
| a |
→ |
c |
(8) |
|
f(g(h(x))) |
→ |
c |
(10) |
|
f(h(g(x))) |
→ |
b |
(3) |
|
f(g(h(x))) |
→ |
b |
(2) |
| a |
→ |
b |
(6) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b] |
= |
1 |
| [f(x1)] |
= |
5 · x1 + 2 |
| [h(x1)] |
= |
1 · x1 + 1 |
| [c] |
= |
0 |
| [g(x1)] |
= |
1 · x1 + 1 |
| [a] |
= |
1 |
all of the following rules can be deleted.
| b |
→ |
c |
(7) |
|
f(h(h(x))) |
→ |
c |
(4) |
|
f(h(g(x))) |
→ |
c |
(9) |
| a |
→ |
c |
(8) |
|
f(g(h(x))) |
→ |
c |
(10) |
|
f(h(g(x))) |
→ |
b |
(3) |
|
f(g(h(x))) |
→ |
b |
(2) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
all of the following rules can be deleted.
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.