Certification Problem
Input (TPDB TRS_Standard/Strategy_removed_CSR_05/Ex3_2_Luc97)
The rewrite relation of the following TRS is considered.
|
dbl(0) |
→ |
0 |
(1) |
|
dbl(s(X)) |
→ |
s(s(dbl(X))) |
(2) |
|
dbls(nil) |
→ |
nil |
(3) |
|
dbls(cons(X,Y)) |
→ |
cons(dbl(X),dbls(Y)) |
(4) |
|
sel(0,cons(X,Y)) |
→ |
X |
(5) |
|
sel(s(X),cons(Y,Z)) |
→ |
sel(X,Z) |
(6) |
|
indx(nil,X) |
→ |
nil |
(7) |
|
indx(cons(X,Y),Z) |
→ |
cons(sel(X,Z),indx(Y,Z)) |
(8) |
|
from(X) |
→ |
cons(X,from(s(X))) |
(9) |
Property / Task
Prove or disprove termination.Answer / Result
No.Proof (by AProVE @ termCOMP 2023)
1 Innermost Lhss Increase
We add the following left hand sides to the innermost strategy.
|
dbl(0) |
|
dbl(s(x0)) |
|
dbls(nil) |
|
dbls(cons(x0,x1)) |
|
sel(0,cons(x0,x1)) |
|
sel(s(x0),cons(x1,x2)) |
|
indx(nil,x0) |
|
indx(cons(x0,x1),x2) |
|
from(x0) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
dbl#(s(X)) |
→ |
dbl#(X) |
(10) |
|
dbls#(cons(X,Y)) |
→ |
dbl#(X) |
(11) |
|
dbls#(cons(X,Y)) |
→ |
dbls#(Y) |
(12) |
|
sel#(s(X),cons(Y,Z)) |
→ |
sel#(X,Z) |
(13) |
|
indx#(cons(X,Y),Z) |
→ |
sel#(X,Z) |
(14) |
|
indx#(cons(X,Y),Z) |
→ |
indx#(Y,Z) |
(15) |
|
from#(X) |
→ |
from#(s(X)) |
(16) |
It remains to prove infiniteness of the resulting DP problem.
1.1.1 Pair and Rule Removal
Some pairs and rules have been removed and it remains to prove infiniteness of the remaing problem.
The following pairs have been deleted.
|
dbl#(s(X)) |
→ |
dbl#(X) |
(10) |
|
dbls#(cons(X,Y)) |
→ |
dbl#(X) |
(11) |
|
dbls#(cons(X,Y)) |
→ |
dbls#(Y) |
(12) |
|
sel#(s(X),cons(Y,Z)) |
→ |
sel#(X,Z) |
(13) |
|
indx#(cons(X,Y),Z) |
→ |
sel#(X,Z) |
(14) |
|
indx#(cons(X,Y),Z) |
→ |
indx#(Y,Z) |
(15) |
and the following rules have been deleted.
1.1.1.1 Pair and Rule Removal
Some pairs and rules have been removed and it remains to prove infiniteness of the remaing problem.
The following pairs have been deleted.
and the following rules have been deleted.
|
dbl(0) |
→ |
0 |
(1) |
|
dbl(s(X)) |
→ |
s(s(dbl(X))) |
(2) |
|
dbls(nil) |
→ |
nil |
(3) |
|
dbls(cons(X,Y)) |
→ |
cons(dbl(X),dbls(Y)) |
(4) |
|
sel(0,cons(X,Y)) |
→ |
X |
(5) |
|
sel(s(X),cons(Y,Z)) |
→ |
sel(X,Z) |
(6) |
|
indx(nil,X) |
→ |
nil |
(7) |
|
indx(cons(X,Y),Z) |
→ |
cons(sel(X,Z),indx(Y,Z)) |
(8) |
|
from(X) |
→ |
cons(X,from(s(X))) |
(9) |
1.1.1.1.1 Innermost Lhss Removal Processor
We restrict the innermost strategy to the following left hand sides.
There are no lhss.
1.1.1.1.1.1 Instantiation Processor
The pairs are instantiated to the following pairs.
|
from#(s(z0)) |
→ |
from#(s(s(z0))) |
(17) |
1.1.1.1.1.1.1 Instantiation Processor
The pairs are instantiated to the following pairs.
|
from#(s(s(z0))) |
→ |
from#(s(s(s(z0)))) |
(18) |
1.1.1.1.1.1.1.1 Loop
The following loop proves infiniteness of the DP problem.
| t0
|
= |
from#(s(s(z0))) |
|
→P
|
from#(s(s(s(z0)))) |
|
= |
t1
|
where t1 =
t0σ
and
σ =
{z0/s(z0)}