Certification Problem
Input (TPDB TRS_Outermost/Strategy_outermost_added_08/Ex5_7_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) |
|
dbl1(0) |
→ |
01 |
(10) |
|
dbl1(s(X)) |
→ |
s1(s1(dbl1(X))) |
(11) |
|
sel1(0,cons(X,Y)) |
→ |
X |
(12) |
|
sel1(s(X),cons(Y,Z)) |
→ |
sel1(X,Z) |
(13) |
|
quote(0) |
→ |
01 |
(14) |
|
quote(s(X)) |
→ |
s1(quote(X)) |
(15) |
|
quote(dbl(X)) |
→ |
dbl1(X) |
(16) |
|
quote(sel(X,Y)) |
→ |
sel1(X,Y) |
(17) |
The evaluation strategy is outermostProperty / Task
Prove or disprove termination.Answer / Result
No.Proof (by AProVE @ termCOMP 2023)
1 Loop
The following loop proves nontermination.
| t0
|
= |
from(X) |
|
→
|
cons(X,from(s(X))) |
|
= |
t1
|
where t1 =
C[t0σ]
and
σ =
{X/s(X)}
and
C = cons(X,☐)