Certification Problem
Input (TPDB TRS_Standard/SK90/4.25)
The rewrite relation of the following TRS is considered.
|
rev(a) |
→ |
a |
(1) |
|
rev(b) |
→ |
b |
(2) |
|
rev(++(x,y)) |
→ |
++(rev(y),rev(x)) |
(3) |
|
rev(++(x,x)) |
→ |
rev(x) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(a) |
= |
0 |
|
weight(a) |
= |
1 |
|
|
|
| prec(b) |
= |
1 |
|
weight(b) |
= |
1 |
|
|
|
| prec(rev) |
= |
3 |
|
weight(rev) |
= |
0 |
|
|
|
| prec(++) |
= |
2 |
|
weight(++) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
rev(a) |
→ |
a |
(1) |
|
rev(b) |
→ |
b |
(2) |
|
rev(++(x,y)) |
→ |
++(rev(y),rev(x)) |
(3) |
|
rev(++(x,x)) |
→ |
rev(x) |
(4) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.