Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex7_BLR02_Z)
The rewrite relation of the following TRS is considered.
from(X) |
→ |
cons(X,n__from(s(X))) |
(1) |
head(cons(X,XS)) |
→ |
X |
(2) |
2nd(cons(X,XS)) |
→ |
head(activate(XS)) |
(3) |
take(0,XS) |
→ |
nil |
(4) |
take(s(N),cons(X,XS)) |
→ |
cons(X,n__take(N,activate(XS))) |
(5) |
sel(0,cons(X,XS)) |
→ |
X |
(6) |
sel(s(N),cons(X,XS)) |
→ |
sel(N,activate(XS)) |
(7) |
from(X) |
→ |
n__from(X) |
(8) |
take(X1,X2) |
→ |
n__take(X1,X2) |
(9) |
activate(n__from(X)) |
→ |
from(X) |
(10) |
activate(n__take(X1,X2)) |
→ |
take(X1,X2) |
(11) |
activate(X) |
→ |
X |
(12) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
prec(from) |
= |
4 |
|
stat(from) |
= |
mul
|
prec(cons) |
= |
1 |
|
stat(cons) |
= |
mul
|
prec(s) |
= |
2 |
|
stat(s) |
= |
mul
|
prec(head) |
= |
5 |
|
stat(head) |
= |
mul
|
prec(2nd) |
= |
6 |
|
stat(2nd) |
= |
mul
|
prec(activate) |
= |
4 |
|
stat(activate) |
= |
mul
|
prec(take) |
= |
4 |
|
stat(take) |
= |
mul
|
prec(0) |
= |
3 |
|
stat(0) |
= |
mul
|
prec(nil) |
= |
3 |
|
stat(nil) |
= |
mul
|
prec(n__take) |
= |
0 |
|
stat(n__take) |
= |
mul
|
prec(sel) |
= |
7 |
|
stat(sel) |
= |
lex
|
π(from) |
= |
[1] |
π(cons) |
= |
[1,2] |
π(n__from) |
= |
1 |
π(s) |
= |
[1] |
π(head) |
= |
[1] |
π(2nd) |
= |
[1] |
π(activate) |
= |
[1] |
π(take) |
= |
[1,2] |
π(0) |
= |
[] |
π(nil) |
= |
[] |
π(n__take) |
= |
[1,2] |
π(sel) |
= |
[1,2] |
all of the following rules can be deleted.
from(X) |
→ |
cons(X,n__from(s(X))) |
(1) |
head(cons(X,XS)) |
→ |
X |
(2) |
2nd(cons(X,XS)) |
→ |
head(activate(XS)) |
(3) |
take(0,XS) |
→ |
nil |
(4) |
take(s(N),cons(X,XS)) |
→ |
cons(X,n__take(N,activate(XS))) |
(5) |
sel(0,cons(X,XS)) |
→ |
X |
(6) |
sel(s(N),cons(X,XS)) |
→ |
sel(N,activate(XS)) |
(7) |
from(X) |
→ |
n__from(X) |
(8) |
take(X1,X2) |
→ |
n__take(X1,X2) |
(9) |
activate(n__take(X1,X2)) |
→ |
take(X1,X2) |
(11) |
activate(X) |
→ |
X |
(12) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(activate) |
= |
2 |
|
weight(activate) |
= |
1 |
|
|
|
prec(n__from) |
= |
0 |
|
weight(n__from) |
= |
1 |
|
|
|
prec(from) |
= |
1 |
|
weight(from) |
= |
2 |
|
|
|
all of the following rules can be deleted.
activate(n__from(X)) |
→ |
from(X) |
(10) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.