Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/PEANO_nosorts-noand_FR)
The rewrite relation of the following TRS is considered.
|
U11(tt,M,N) |
→ |
U12(tt,activate(M),activate(N)) |
(1) |
|
U12(tt,M,N) |
→ |
s(plus(activate(N),activate(M))) |
(2) |
|
plus(N,0) |
→ |
N |
(3) |
|
plus(N,s(M)) |
→ |
U11(tt,M,N) |
(4) |
|
activate(X) |
→ |
X |
(5) |
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
| [U11(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 2 · x3
|
| [tt] |
= |
0 |
| [U12(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 2 · x3
|
| [activate(x1)] |
= |
1 · x1
|
| [s(x1)] |
= |
1 · x1
|
| [plus(x1, x2)] |
= |
2 · x1 + 1 · x2
|
| [0] |
= |
2 |
all of the following rules can be deleted.
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(tt) |
= |
3 |
|
weight(tt) |
= |
2 |
|
|
|
| prec(activate) |
= |
5 |
|
weight(activate) |
= |
0 |
|
|
|
| prec(s) |
= |
0 |
|
weight(s) |
= |
1 |
|
|
|
| prec(U11) |
= |
2 |
|
weight(U11) |
= |
0 |
|
|
|
| prec(U12) |
= |
1 |
|
weight(U12) |
= |
0 |
|
|
|
| prec(plus) |
= |
4 |
|
weight(plus) |
= |
1 |
|
|
|
all of the following rules can be deleted.
|
U11(tt,M,N) |
→ |
U12(tt,activate(M),activate(N)) |
(1) |
|
U12(tt,M,N) |
→ |
s(plus(activate(N),activate(M))) |
(2) |
|
plus(N,s(M)) |
→ |
U11(tt,M,N) |
(4) |
|
activate(X) |
→ |
X |
(5) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.