Certification Problem
Input (COPS 960)
We consider the TRS containing the following rules:
|
a(x) |
→ |
x |
(1) |
|
a(b(x)) |
→ |
c(b(b(a(a(x))))) |
(2) |
|
b(x) |
→ |
c(x) |
(3) |
|
c(c(x)) |
→ |
x |
(4) |
The underlying signature is as follows:
{a/1, b/1, c/1}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:
|
c(c(x)) |
→ |
x |
(4) |
|
b(x) |
→ |
c(x) |
(3) |
|
a(b(x)) |
→ |
c(b(b(a(a(x))))) |
(2) |
|
a(x) |
→ |
x |
(1) |
|
a(b(x)) |
→ |
c(b(c(a(a(x))))) |
(5) |
|
a(b(x)) |
→ |
c(c(b(a(a(x))))) |
(6) |
|
a(b(x)) |
→ |
c(b(b(a(x)))) |
(7) |
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:
|
a(x) |
→ |
x |
(1) |
|
b(x) |
→ |
c(x) |
(3) |
|
c(c(x)) |
→ |
x |
(4) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b(x1)] |
= |
1 · x1 + 0 |
| [a(x1)] |
= |
1 · x1 + 1 |
| [c(x1)] |
= |
1 · x1 + 0 |
all of the following rules can be deleted.
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b(x1)] |
= |
4 · x1 + 4 |
| [c(x1)] |
= |
2 · x1 + 4 |
all of the following rules can be deleted.
1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b(x1)] |
= |
4 · x1 + 1 |
| [c(x1)] |
= |
4 · x1 + 0 |
all of the following rules can be deleted.
1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.