Certification Problem
Input (COPS 190)
We consider the TRS containing the following rules:
|
or(x,T) |
→ |
T |
(1) |
|
or(x,F) |
→ |
x |
(2) |
|
or(x,y) |
→ |
or(y,x) |
(3) |
The underlying signature is as follows:
{or/2, T/0, F/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:
|
or(x,y) |
→ |
or(y,x) |
(3) |
|
or(x,F) |
→ |
x |
(2) |
|
or(x,T) |
→ |
T |
(1) |
|
or(T,x) |
→ |
T |
(4) |
|
or(F,x) |
→ |
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:
|
or(F,x) |
→ |
x |
(5) |
|
or(T,x) |
→ |
T |
(4) |
|
or(x,T) |
→ |
T |
(1) |
|
or(x,F) |
→ |
x |
(2) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [or(x1, x2)] |
= |
4 · x1 + 1 · x2 + 0 |
| [T] |
= |
0 |
| [F] |
= |
4 |
all of the following rules can be deleted.
|
or(F,x) |
→ |
x |
(5) |
|
or(x,F) |
→ |
x |
(2) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [or(x1, x2)] |
= |
4 · x1 + 2 · x2 + 6 |
| [T] |
= |
5 |
all of the following rules can be deleted.
|
or(T,x) |
→ |
T |
(4) |
|
or(x,T) |
→ |
T |
(1) |
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.