Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z106)
The rewrite relation of the following TRS is considered.
|
a(a(x1)) |
→ |
b(b(b(x1))) |
(1) |
|
b(x1) |
→ |
c(c(d(x1))) |
(2) |
|
c(x1) |
→ |
d(d(d(x1))) |
(3) |
|
b(c(x1)) |
→ |
c(b(x1)) |
(4) |
|
b(c(d(x1))) |
→ |
a(x1) |
(5) |
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
|
a(a(x1)) |
→ |
b(b(b(x1))) |
(1) |
|
b(x1) |
→ |
d(c(c(x1))) |
(6) |
|
c(x1) |
→ |
d(d(d(x1))) |
(3) |
|
c(b(x1)) |
→ |
b(c(x1)) |
(7) |
|
d(c(b(x1))) |
→ |
a(x1) |
(8) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [b(x1)] |
= |
2 · x1 +
-∞ |
| [c(x1)] |
= |
1 · x1 +
-∞ |
| [a(x1)] |
= |
3 · x1 +
-∞ |
| [d(x1)] |
= |
0 · x1 +
-∞ |
all of the following rules can be deleted.
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(c) |
= |
3 |
|
weight(c) |
= |
0 |
|
|
|
| prec(d) |
= |
1 |
|
weight(d) |
= |
2 |
|
|
|
| prec(b) |
= |
2 |
|
weight(b) |
= |
2 |
|
|
|
| prec(a) |
= |
0 |
|
weight(a) |
= |
4 |
|
|
|
all of the following rules can be deleted.
|
a(a(x1)) |
→ |
b(b(b(x1))) |
(1) |
|
b(x1) |
→ |
d(c(c(x1))) |
(6) |
|
c(b(x1)) |
→ |
b(c(x1)) |
(7) |
|
d(c(b(x1))) |
→ |
a(x1) |
(8) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.