Certification Problem
Input (TPDB SRS_Standard/Mixed_SRS/turing_add)
The rewrite relation of the following TRS is considered.
|
1(q0(1(x1))) |
→ |
0(1(q1(x1))) |
(1) |
|
1(q0(0(x1))) |
→ |
0(0(q1(x1))) |
(2) |
|
1(q1(1(x1))) |
→ |
1(1(q1(x1))) |
(3) |
|
1(q1(0(x1))) |
→ |
1(0(q1(x1))) |
(4) |
|
0(q1(x1)) |
→ |
q2(1(x1)) |
(5) |
|
1(q2(x1)) |
→ |
q2(1(x1)) |
(6) |
|
0(q2(x1)) |
→ |
0(q0(x1)) |
(7) |
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(q2) |
= |
1 |
|
weight(q2) |
= |
5 |
|
|
|
| prec(0) |
= |
2 |
|
weight(0) |
= |
6 |
|
|
|
| prec(q1) |
= |
4 |
|
weight(q1) |
= |
3 |
|
|
|
| prec(q0) |
= |
0 |
|
weight(q0) |
= |
5 |
|
|
|
| prec(1) |
= |
3 |
|
weight(1) |
= |
4 |
|
|
|
all of the following rules can be deleted.
|
1(q0(1(x1))) |
→ |
0(1(q1(x1))) |
(1) |
|
1(q0(0(x1))) |
→ |
0(0(q1(x1))) |
(2) |
|
1(q1(1(x1))) |
→ |
1(1(q1(x1))) |
(3) |
|
1(q1(0(x1))) |
→ |
1(0(q1(x1))) |
(4) |
|
0(q1(x1)) |
→ |
q2(1(x1)) |
(5) |
|
1(q2(x1)) |
→ |
q2(1(x1)) |
(6) |
|
0(q2(x1)) |
→ |
0(q0(x1)) |
(7) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.