Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/25192)
The rewrite relation of the following TRS is considered.
|
0(x1) |
→ |
1(x1) |
(1) |
|
4(5(4(5(x1)))) |
→ |
4(4(5(5(x1)))) |
(2) |
|
5(5(5(5(5(5(4(4(4(4(4(4(x1)))))))))))) |
→ |
2(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
|
0(x1) |
→ |
1(x1) |
(1) |
|
5(4(5(4(x1)))) |
→ |
5(5(4(4(x1)))) |
(4) |
|
4(4(4(4(4(4(5(5(5(5(5(5(x1)))))))))))) |
→ |
2(x1) |
(5) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(2) |
= |
2 |
|
weight(2) |
= |
1 |
|
|
|
| prec(4) |
= |
4 |
|
weight(4) |
= |
0 |
|
|
|
| prec(5) |
= |
1 |
|
weight(5) |
= |
1 |
|
|
|
| prec(1) |
= |
0 |
|
weight(1) |
= |
1 |
|
|
|
| prec(0) |
= |
3 |
|
weight(0) |
= |
5 |
|
|
|
all of the following rules can be deleted.
|
0(x1) |
→ |
1(x1) |
(1) |
|
5(4(5(4(x1)))) |
→ |
5(5(4(4(x1)))) |
(4) |
|
4(4(4(4(4(4(5(5(5(5(5(5(x1)))))))))))) |
→ |
2(x1) |
(5) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.