Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z113)
The rewrite relation of the following TRS is considered.
1(1(x1)) |
→ |
4(3(x1)) |
(1) |
1(2(x1)) |
→ |
2(1(x1)) |
(2) |
2(2(x1)) |
→ |
1(1(1(x1))) |
(3) |
3(3(x1)) |
→ |
5(6(x1)) |
(4) |
3(4(x1)) |
→ |
1(1(x1)) |
(5) |
4(4(x1)) |
→ |
3(x1) |
(6) |
5(5(x1)) |
→ |
6(2(x1)) |
(7) |
5(6(x1)) |
→ |
1(2(x1)) |
(8) |
6(6(x1)) |
→ |
2(1(x1)) |
(9) |
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(1(x1)) |
→ |
3(4(x1)) |
(10) |
2(1(x1)) |
→ |
1(2(x1)) |
(11) |
2(2(x1)) |
→ |
1(1(1(x1))) |
(3) |
3(3(x1)) |
→ |
6(5(x1)) |
(12) |
4(3(x1)) |
→ |
1(1(x1)) |
(13) |
4(4(x1)) |
→ |
3(x1) |
(6) |
5(5(x1)) |
→ |
2(6(x1)) |
(14) |
6(5(x1)) |
→ |
2(1(x1)) |
(15) |
6(6(x1)) |
→ |
1(2(x1)) |
(16) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[6(x1)] |
= |
1 · x1 + 15 |
[3(x1)] |
= |
1 · x1 + 16 |
[2(x1)] |
= |
1 · x1 + 18 |
[5(x1)] |
= |
1 · x1 + 17 |
[1(x1)] |
= |
1 · x1 + 12 |
[4(x1)] |
= |
1 · x1 + 8 |
all of the following rules can be deleted.
5(5(x1)) |
→ |
2(6(x1)) |
(14) |
6(5(x1)) |
→ |
2(1(x1)) |
(15) |
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(5) |
= |
5 |
|
weight(5) |
= |
0 |
|
|
|
prec(6) |
= |
0 |
|
weight(6) |
= |
4 |
|
|
|
prec(2) |
= |
3 |
|
weight(2) |
= |
3 |
|
|
|
prec(4) |
= |
4 |
|
weight(4) |
= |
2 |
|
|
|
prec(3) |
= |
1 |
|
weight(3) |
= |
2 |
|
|
|
prec(1) |
= |
2 |
|
weight(1) |
= |
2 |
|
|
|
all of the following rules can be deleted.
1(1(x1)) |
→ |
3(4(x1)) |
(10) |
2(1(x1)) |
→ |
1(2(x1)) |
(11) |
2(2(x1)) |
→ |
1(1(1(x1))) |
(3) |
3(3(x1)) |
→ |
6(5(x1)) |
(12) |
4(3(x1)) |
→ |
1(1(x1)) |
(13) |
4(4(x1)) |
→ |
3(x1) |
(6) |
6(6(x1)) |
→ |
1(2(x1)) |
(16) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.