Certification Problem
Input (TPDB TRS_Standard/AProVE_04/Liveness6.1)
The rewrite relation of the following TRS is considered.
|
top(free(x)) |
→ |
top(check(new(x))) |
(1) |
|
new(free(x)) |
→ |
free(new(x)) |
(2) |
|
old(free(x)) |
→ |
free(old(x)) |
(3) |
|
new(serve) |
→ |
free(serve) |
(4) |
|
old(serve) |
→ |
free(serve) |
(5) |
|
check(free(x)) |
→ |
free(check(x)) |
(6) |
|
check(new(x)) |
→ |
new(check(x)) |
(7) |
|
check(old(x)) |
→ |
old(check(x)) |
(8) |
|
check(old(x)) |
→ |
old(x) |
(9) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [old(x1)] |
= |
· x1 +
|
| [top(x1)] |
= |
· x1 +
|
| [check(x1)] |
= |
· x1 +
|
| [serve] |
= |
|
| [free(x1)] |
= |
· x1 +
|
| [new(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
old(serve) |
→ |
free(serve) |
(5) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [old(x1)] |
= |
· x1 +
|
| [top(x1)] |
= |
· x1 +
|
| [check(x1)] |
= |
· x1 +
|
| [serve] |
= |
|
| [free(x1)] |
= |
· x1 +
|
| [new(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
top(free(x)) |
→ |
top(check(new(x))) |
(1) |
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(serve) |
= |
6 |
|
weight(serve) |
= |
4 |
|
|
|
| prec(old) |
= |
5 |
|
weight(old) |
= |
2 |
|
|
|
| prec(check) |
= |
7 |
|
weight(check) |
= |
2 |
|
|
|
| prec(new) |
= |
1 |
|
weight(new) |
= |
2 |
|
|
|
| prec(free) |
= |
0 |
|
weight(free) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
new(free(x)) |
→ |
free(new(x)) |
(2) |
|
old(free(x)) |
→ |
free(old(x)) |
(3) |
|
new(serve) |
→ |
free(serve) |
(4) |
|
check(free(x)) |
→ |
free(check(x)) |
(6) |
|
check(new(x)) |
→ |
new(check(x)) |
(7) |
|
check(old(x)) |
→ |
old(check(x)) |
(8) |
|
check(old(x)) |
→ |
old(x) |
(9) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.