Certification Problem
Input (TPDB TRS_Standard/CiME_04/dpqs)
The rewrite relation of the following TRS is considered.
|
f(f(x)) |
→ |
f(c(f(x))) |
(1) |
|
f(f(x)) |
→ |
f(d(f(x))) |
(2) |
|
g(c(x)) |
→ |
x |
(3) |
|
g(d(x)) |
→ |
x |
(4) |
|
g(c(0)) |
→ |
g(d(1)) |
(5) |
|
g(c(1)) |
→ |
g(d(0)) |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [0] |
= |
|
| [c(x1)] |
= |
· x1 +
|
| [g(x1)] |
= |
· x1 +
|
| [1] |
= |
|
| [f(x1)] |
= |
· x1 +
|
| [d(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
g(c(x)) |
→ |
x |
(3) |
|
g(d(x)) |
→ |
x |
(4) |
|
g(c(0)) |
→ |
g(d(1)) |
(5) |
|
g(c(1)) |
→ |
g(d(0)) |
(6) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [c(x1)] |
= |
· x1 +
|
| [f(x1)] |
= |
· x1 +
|
| [d(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1
over the naturals
| [c(x1)] |
= |
|
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
1 |
0 |
0 |
| 0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
|
| [f(x1)] |
= |
|
| 1 |
0 |
0 |
1 |
1 |
| 0 |
0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
1 |
0 |
| 0 |
0 |
1 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.