Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-152)
The rewrite relation of the following TRS is considered.
|
b(b(b(b(x1)))) |
→ |
b(b(a(a(x1)))) |
(1) |
|
b(b(a(a(x1)))) |
→ |
a(b(a(b(x1)))) |
(2) |
|
a(b(a(a(x1)))) |
→ |
a(b(a(b(x1)))) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
b(b(b(b(x1)))) |
→ |
a(a(b(b(x1)))) |
(4) |
|
a(a(b(b(x1)))) |
→ |
b(a(b(a(x1)))) |
(5) |
|
a(a(b(a(x1)))) |
→ |
b(a(b(a(x1)))) |
(6) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1
over the naturals
| [a(x1)] |
= |
|
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
1 |
0 |
0 |
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 |
|
|
|
| [b(x1)] |
= |
|
| 1 |
0 |
1 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
1 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
1 |
0 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
|
b(b(b(b(x1)))) |
→ |
a(a(b(b(x1)))) |
(4) |
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(a) |
= |
1 |
|
weight(a) |
= |
2 |
|
|
|
| prec(b) |
= |
0 |
|
weight(b) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
a(a(b(b(x1)))) |
→ |
b(a(b(a(x1)))) |
(5) |
|
a(a(b(a(x1)))) |
→ |
b(a(b(a(x1)))) |
(6) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.