Certification Problem
Input (TPDB SRS_Standard/Zantema_06/beans1)
The rewrite relation of the following TRS is considered.
|
1(2(1(x1))) |
→ |
2(0(2(x1))) |
(1) |
|
0(2(1(x1))) |
→ |
1(0(2(x1))) |
(2) |
|
L(2(1(x1))) |
→ |
L(1(0(2(x1)))) |
(3) |
|
1(2(0(x1))) |
→ |
2(0(1(x1))) |
(4) |
|
1(2(R(x1))) |
→ |
2(0(1(R(x1)))) |
(5) |
|
0(2(0(x1))) |
→ |
1(0(1(x1))) |
(6) |
|
L(2(0(x1))) |
→ |
L(1(0(1(x1)))) |
(7) |
|
0(2(R(x1))) |
→ |
1(0(1(R(x1)))) |
(8) |
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
|
1(2(1(x1))) |
→ |
2(0(2(x1))) |
(1) |
|
1(2(0(x1))) |
→ |
2(0(1(x1))) |
(4) |
|
1(2(L(x1))) |
→ |
2(0(1(L(x1)))) |
(9) |
|
0(2(1(x1))) |
→ |
1(0(2(x1))) |
(2) |
|
R(2(1(x1))) |
→ |
R(1(0(2(x1)))) |
(10) |
|
0(2(0(x1))) |
→ |
1(0(1(x1))) |
(6) |
|
0(2(L(x1))) |
→ |
1(0(1(L(x1)))) |
(11) |
|
R(2(0(x1))) |
→ |
R(1(0(1(x1)))) |
(12) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [R(x1)] |
= |
· x1 +
|
| [2(x1)] |
= |
· x1 +
|
| [L(x1)] |
= |
· x1 +
|
| [1(x1)] |
= |
· x1 +
|
| [0(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
1(2(L(x1))) |
→ |
2(0(1(L(x1)))) |
(9) |
|
0(2(L(x1))) |
→ |
1(0(1(L(x1)))) |
(11) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [R(x1)] |
= |
· x1 +
|
| [2(x1)] |
= |
· x1 +
|
| [1(x1)] |
= |
· x1 +
|
| [0(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
R(2(1(x1))) |
→ |
R(1(0(2(x1)))) |
(10) |
|
R(2(0(x1))) |
→ |
R(1(0(1(x1)))) |
(12) |
1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(0) |
= |
3 |
|
weight(0) |
= |
0 |
|
|
|
| prec(2) |
= |
0 |
|
weight(2) |
= |
2 |
|
|
|
| prec(1) |
= |
2 |
|
weight(1) |
= |
1 |
|
|
|
all of the following rules can be deleted.
|
1(2(1(x1))) |
→ |
2(0(2(x1))) |
(1) |
|
1(2(0(x1))) |
→ |
2(0(1(x1))) |
(4) |
|
0(2(1(x1))) |
→ |
1(0(2(x1))) |
(2) |
|
0(2(0(x1))) |
→ |
1(0(1(x1))) |
(6) |
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.