Certification Problem
Input (TPDB TRS_Standard/Mixed_TRS/jones5)
The rewrite relation of the following TRS is considered.
|
f(x,empty) |
→ |
x |
(1) |
|
f(empty,cons(a,k)) |
→ |
f(cons(a,k),k) |
(2) |
|
f(cons(a,k),y) |
→ |
f(y,k) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1, x2)] |
= |
1 · x1 + 2 · x2
|
| [empty] |
= |
2 |
| [cons(x1, x2)] |
= |
1 · x1 + 2 · x2
|
all of the following rules can be deleted.
|
f(x,empty) |
→ |
x |
(1) |
|
f(empty,cons(a,k)) |
→ |
f(cons(a,k),k) |
(2) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(cons) |
= |
1 |
|
weight(cons) |
= |
0 |
|
|
|
| prec(f) |
= |
0 |
|
weight(f) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
f(cons(a,k),y) |
→ |
f(y,k) |
(3) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.