Certification Problem
Input (TPDB TRS_Standard/Mixed_TRS/jones1)
The rewrite relation of the following TRS is considered.
|
rev(ls) |
→ |
r1(ls,empty) |
(1) |
|
r1(empty,a) |
→ |
a |
(2) |
|
r1(cons(x,k),a) |
→ |
r1(k,cons(x,a)) |
(3) |
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(empty) |
= |
2 |
|
weight(empty) |
= |
1 |
|
|
|
| prec(rev) |
= |
1 |
|
weight(rev) |
= |
1 |
|
|
|
| prec(r1) |
= |
0 |
|
weight(r1) |
= |
0 |
|
|
|
| prec(cons) |
= |
3 |
|
weight(cons) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
rev(ls) |
→ |
r1(ls,empty) |
(1) |
|
r1(empty,a) |
→ |
a |
(2) |
|
r1(cons(x,k),a) |
→ |
r1(k,cons(x,a)) |
(3) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.