Certification Problem
Input (TPDB TRS_Standard/SK90/2.48)
The rewrite relation of the following TRS is considered.
|
d(x) |
→ |
e(u(x)) |
(1) |
|
d(u(x)) |
→ |
c(x) |
(2) |
|
c(u(x)) |
→ |
b(x) |
(3) |
|
v(e(x)) |
→ |
x |
(4) |
|
b(u(x)) |
→ |
a(e(x)) |
(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
|
d(x) |
→ |
u(e(x)) |
(6) |
|
u(d(x)) |
→ |
c(x) |
(7) |
|
u(c(x)) |
→ |
b(x) |
(8) |
|
e(v(x)) |
→ |
x |
(9) |
|
u(b(x)) |
→ |
e(a(x)) |
(10) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(a) |
= |
7 |
|
weight(a) |
= |
3 |
|
|
|
| prec(v) |
= |
5 |
|
weight(v) |
= |
2 |
|
|
|
| prec(b) |
= |
0 |
|
weight(b) |
= |
4 |
|
|
|
| prec(c) |
= |
1 |
|
weight(c) |
= |
3 |
|
|
|
| prec(e) |
= |
4 |
|
weight(e) |
= |
1 |
|
|
|
| prec(u) |
= |
2 |
|
weight(u) |
= |
1 |
|
|
|
| prec(d) |
= |
3 |
|
weight(d) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
d(x) |
→ |
u(e(x)) |
(6) |
|
u(d(x)) |
→ |
c(x) |
(7) |
|
u(c(x)) |
→ |
b(x) |
(8) |
|
e(v(x)) |
→ |
x |
(9) |
|
u(b(x)) |
→ |
e(a(x)) |
(10) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.