Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex4_7_56_Bor03_FR)
The rewrite relation of the following TRS is considered.
from(X) |
→ |
cons(X,n__from(n__s(X))) |
(1) |
after(0,XS) |
→ |
XS |
(2) |
after(s(N),cons(X,XS)) |
→ |
after(N,activate(XS)) |
(3) |
from(X) |
→ |
n__from(X) |
(4) |
s(X) |
→ |
n__s(X) |
(5) |
activate(n__from(X)) |
→ |
from(activate(X)) |
(6) |
activate(n__s(X)) |
→ |
s(activate(X)) |
(7) |
activate(X) |
→ |
X |
(8) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
prec(from) |
= |
3 |
|
stat(from) |
= |
mul
|
prec(cons) |
= |
0 |
|
stat(cons) |
= |
mul
|
prec(n__from) |
= |
1 |
|
stat(n__from) |
= |
mul
|
prec(n__s) |
= |
2 |
|
stat(n__s) |
= |
mul
|
prec(after) |
= |
6 |
|
stat(after) |
= |
lex
|
prec(0) |
= |
7 |
|
stat(0) |
= |
mul
|
prec(s) |
= |
4 |
|
stat(s) |
= |
mul
|
prec(activate) |
= |
5 |
|
stat(activate) |
= |
mul
|
π(from) |
= |
[1] |
π(cons) |
= |
[1,2] |
π(n__from) |
= |
[1] |
π(n__s) |
= |
[1] |
π(after) |
= |
[1,2] |
π(0) |
= |
[] |
π(s) |
= |
[1] |
π(activate) |
= |
[1] |
all of the following rules can be deleted.
from(X) |
→ |
cons(X,n__from(n__s(X))) |
(1) |
after(0,XS) |
→ |
XS |
(2) |
after(s(N),cons(X,XS)) |
→ |
after(N,activate(XS)) |
(3) |
from(X) |
→ |
n__from(X) |
(4) |
s(X) |
→ |
n__s(X) |
(5) |
activate(n__from(X)) |
→ |
from(activate(X)) |
(6) |
activate(n__s(X)) |
→ |
s(activate(X)) |
(7) |
activate(X) |
→ |
X |
(8) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.