Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z092)
The rewrite relation of the following TRS is considered.
|
q0(0(x1)) |
→ |
0'(q1(x1)) |
(1) |
|
q1(0(x1)) |
→ |
0(q1(x1)) |
(2) |
|
q1(1'(x1)) |
→ |
1'(q1(x1)) |
(3) |
|
0(q1(1(x1))) |
→ |
q2(0(1'(x1))) |
(4) |
|
0'(q1(1(x1))) |
→ |
q2(0'(1'(x1))) |
(5) |
|
1'(q1(1(x1))) |
→ |
q2(1'(1'(x1))) |
(6) |
|
0(q2(0(x1))) |
→ |
q2(0(0(x1))) |
(7) |
|
0'(q2(0(x1))) |
→ |
q2(0'(0(x1))) |
(8) |
|
1'(q2(0(x1))) |
→ |
q2(1'(0(x1))) |
(9) |
|
0(q2(1'(x1))) |
→ |
q2(0(1'(x1))) |
(10) |
|
0'(q2(1'(x1))) |
→ |
q2(0'(1'(x1))) |
(11) |
|
1'(q2(1'(x1))) |
→ |
q2(1'(1'(x1))) |
(12) |
|
q2(0'(x1)) |
→ |
0'(q0(x1)) |
(13) |
|
q0(1'(x1)) |
→ |
1'(q3(x1)) |
(14) |
|
q3(1'(x1)) |
→ |
1'(q3(x1)) |
(15) |
|
q3(b(x1)) |
→ |
b(q4(x1)) |
(16) |
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(q0(x1)) |
→ |
q1(0'(x1)) |
(17) |
|
0(q1(x1)) |
→ |
q1(0(x1)) |
(18) |
|
1'(q1(x1)) |
→ |
q1(1'(x1)) |
(19) |
|
1(q1(0(x1))) |
→ |
1'(0(q2(x1))) |
(20) |
|
1(q1(0'(x1))) |
→ |
1'(0'(q2(x1))) |
(21) |
|
1(q1(1'(x1))) |
→ |
1'(1'(q2(x1))) |
(22) |
|
0(q2(0(x1))) |
→ |
0(0(q2(x1))) |
(23) |
|
0(q2(0'(x1))) |
→ |
0(0'(q2(x1))) |
(24) |
|
0(q2(1'(x1))) |
→ |
0(1'(q2(x1))) |
(25) |
|
1'(q2(0(x1))) |
→ |
1'(0(q2(x1))) |
(26) |
|
1'(q2(0'(x1))) |
→ |
1'(0'(q2(x1))) |
(27) |
|
1'(q2(1'(x1))) |
→ |
1'(1'(q2(x1))) |
(28) |
|
0'(q2(x1)) |
→ |
q0(0'(x1)) |
(29) |
|
1'(q0(x1)) |
→ |
q3(1'(x1)) |
(30) |
|
1'(q3(x1)) |
→ |
q3(1'(x1)) |
(31) |
|
b(q3(x1)) |
→ |
q4(b(x1)) |
(32) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(q4) |
= |
2 |
|
weight(q4) |
= |
2 |
|
|
|
| prec(b) |
= |
11 |
|
weight(b) |
= |
2 |
|
|
|
| prec(q3) |
= |
4 |
|
weight(q3) |
= |
2 |
|
|
|
| prec(q2) |
= |
7 |
|
weight(q2) |
= |
5 |
|
|
|
| prec(1) |
= |
14 |
|
weight(1) |
= |
5 |
|
|
|
| prec(1') |
= |
5 |
|
weight(1') |
= |
2 |
|
|
|
| prec(0') |
= |
6 |
|
weight(0') |
= |
3 |
|
|
|
| prec(q1) |
= |
0 |
|
weight(q1) |
= |
2 |
|
|
|
| prec(q0) |
= |
15 |
|
weight(q0) |
= |
3 |
|
|
|
| prec(0) |
= |
1 |
|
weight(0) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
0(q0(x1)) |
→ |
q1(0'(x1)) |
(17) |
|
0(q1(x1)) |
→ |
q1(0(x1)) |
(18) |
|
1'(q1(x1)) |
→ |
q1(1'(x1)) |
(19) |
|
1(q1(0(x1))) |
→ |
1'(0(q2(x1))) |
(20) |
|
1(q1(0'(x1))) |
→ |
1'(0'(q2(x1))) |
(21) |
|
1(q1(1'(x1))) |
→ |
1'(1'(q2(x1))) |
(22) |
|
0(q2(0(x1))) |
→ |
0(0(q2(x1))) |
(23) |
|
0(q2(0'(x1))) |
→ |
0(0'(q2(x1))) |
(24) |
|
0(q2(1'(x1))) |
→ |
0(1'(q2(x1))) |
(25) |
|
1'(q2(0(x1))) |
→ |
1'(0(q2(x1))) |
(26) |
|
1'(q2(0'(x1))) |
→ |
1'(0'(q2(x1))) |
(27) |
|
1'(q2(1'(x1))) |
→ |
1'(1'(q2(x1))) |
(28) |
|
0'(q2(x1)) |
→ |
q0(0'(x1)) |
(29) |
|
1'(q0(x1)) |
→ |
q3(1'(x1)) |
(30) |
|
1'(q3(x1)) |
→ |
q3(1'(x1)) |
(31) |
|
b(q3(x1)) |
→ |
q4(b(x1)) |
(32) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.