Certification Problem
Input (COPS 944)
We consider the TRS containing the following rules:
|
C(x) |
→ |
c(x) |
(1) |
|
c(c(x)) |
→ |
x |
(2) |
|
b(b(x)) |
→ |
B(x) |
(3) |
|
B(B(x)) |
→ |
b(x) |
(4) |
|
c(B(c(b(c(x))))) |
→ |
B(c(b(c(B(c(b(x))))))) |
(5) |
|
b(B(x)) |
→ |
x |
(6) |
|
B(b(x)) |
→ |
x |
(7) |
|
c(C(x)) |
→ |
x |
(8) |
|
C(c(x)) |
→ |
x |
(9) |
The underlying signature is as follows:
{C/1, c/1, b/1, B/1}Property / Task
Prove or disprove confluence.Answer / Result
Yes.Proof (by csi @ CoCo 2022)
1 Locally confluent and terminating
Confluence is proven by showing local confluence and termination.
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [c(x1)] |
= |
2 · x1 + 1 |
| [B(x1)] |
= |
1 · x1 + 0 |
| [C(x1)] |
= |
2 · x1 + 1 |
| [b(x1)] |
= |
1 · x1 + 0 |
all of the following rules can be deleted.
|
c(c(x)) |
→ |
x |
(2) |
|
c(C(x)) |
→ |
x |
(8) |
|
C(c(x)) |
→ |
x |
(9) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [c(x1)] |
= |
1 · x1 + 0 |
| [B(x1)] |
= |
1 · x1 + 0 |
| [C(x1)] |
= |
2 · x1 + 4 |
| [b(x1)] |
= |
1 · x1 + 0 |
all of the following rules can be deleted.
1.1.1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
b(b(x)) |
→ |
B(x) |
(3) |
|
B(B(x)) |
→ |
b(x) |
(4) |
|
c(b(c(B(c(x))))) |
→ |
b(c(B(c(b(c(B(x))))))) |
(10) |
|
B(b(x)) |
→ |
x |
(7) |
|
b(B(x)) |
→ |
x |
(6) |
1.1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(b(x)) |
→ |
B#(x) |
(11) |
|
B#(B(x)) |
→ |
b#(x) |
(12) |
|
c#(b(c(B(c(x))))) |
→ |
B#(x) |
(13) |
|
c#(b(c(B(c(x))))) |
→ |
c#(B(x)) |
(14) |
|
c#(b(c(B(c(x))))) |
→ |
b#(c(B(x))) |
(15) |
|
c#(b(c(B(c(x))))) |
→ |
c#(b(c(B(x)))) |
(16) |
|
c#(b(c(B(c(x))))) |
→ |
B#(c(b(c(B(x))))) |
(17) |
|
c#(b(c(B(c(x))))) |
→ |
c#(B(c(b(c(B(x)))))) |
(18) |
|
c#(b(c(B(c(x))))) |
→ |
b#(c(B(c(b(c(B(x))))))) |
(19) |
1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
1.2 Local Confluence Proof
All critical pairs are joinable which can be seen by computing normal forms of all critical pairs.