Certification Problem
Input (TPDB TRS_Outermost/Strategy_outermost_added_08/Ex4_7_37_Bor03_L)
The rewrite relation of the following TRS is considered.
|
sel(s(N),cons(X)) |
→ |
sel(N,XS) |
(1) |
|
from(X) |
→ |
cons(X) |
(2) |
|
sel(0,cons(X)) |
→ |
X |
(3) |
|
minus(X,0) |
→ |
0 |
(4) |
|
minus(s(X),s(Y)) |
→ |
minus(X,Y) |
(5) |
|
quot(0,s(Y)) |
→ |
0 |
(6) |
|
quot(s(X),s(Y)) |
→ |
s(quot(minus(X,Y),s(Y))) |
(7) |
|
zWquot(XS,nil) |
→ |
nil |
(8) |
|
zWquot(nil,XS) |
→ |
nil |
(9) |
|
zWquot(cons(X),cons(Y)) |
→ |
cons(quot(X,Y)) |
(10) |
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
|
= |
sel(s(N),cons(X)) |
|
→
|
sel(N,sel(s(N),cons(X))) |
|
= |
t1
|
where t1 =
C[t0]
and
C = sel(N,☐)