Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/264405)
The rewrite relation of the following TRS is considered.
|
0(1(2(3(4(x1))))) |
→ |
0(2(3(1(4(x1))))) |
(1) |
|
0(5(1(2(3(4(x1)))))) |
→ |
0(1(2(5(3(4(x1)))))) |
(2) |
|
0(5(1(2(3(4(x1)))))) |
→ |
0(5(2(1(3(4(x1)))))) |
(3) |
|
0(5(1(2(3(4(x1)))))) |
→ |
5(0(2(3(1(4(x1)))))) |
(4) |
|
0(5(2(3(1(4(x1)))))) |
→ |
0(1(5(2(3(4(x1)))))) |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(5) |
= |
2 |
|
weight(5) |
= |
2 |
|
|
|
| prec(0) |
= |
7 |
|
weight(0) |
= |
2 |
|
|
|
| prec(1) |
= |
1 |
|
weight(1) |
= |
2 |
|
|
|
| prec(2) |
= |
0 |
|
weight(2) |
= |
2 |
|
|
|
| prec(3) |
= |
6 |
|
weight(3) |
= |
2 |
|
|
|
| prec(4) |
= |
4 |
|
weight(4) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
0(1(2(3(4(x1))))) |
→ |
0(2(3(1(4(x1))))) |
(1) |
|
0(5(1(2(3(4(x1)))))) |
→ |
0(1(2(5(3(4(x1)))))) |
(2) |
|
0(5(1(2(3(4(x1)))))) |
→ |
0(5(2(1(3(4(x1)))))) |
(3) |
|
0(5(1(2(3(4(x1)))))) |
→ |
5(0(2(3(1(4(x1)))))) |
(4) |
|
0(5(2(3(1(4(x1)))))) |
→ |
0(1(5(2(3(4(x1)))))) |
(5) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.