Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex5_7_Luc97_L)
The rewrite relation of the following TRS is considered.
|
sel(0,cons) |
→ |
X |
(1) |
|
sel(s,cons) |
→ |
sel(X,Z) |
(2) |
|
dbl1(s) |
→ |
s1(s1(dbl1(X))) |
(3) |
|
sel1(0,cons) |
→ |
X |
(4) |
|
sel1(s,cons) |
→ |
sel1(X,Z) |
(5) |
|
quote(s) |
→ |
s1(quote(X)) |
(6) |
|
dbl(0) |
→ |
0 |
(7) |
|
dbl(s) |
→ |
s |
(8) |
|
dbls(nil) |
→ |
nil |
(9) |
|
dbls(cons) |
→ |
cons |
(10) |
|
indx(nil) |
→ |
nil |
(11) |
|
indx(cons) |
→ |
cons |
(12) |
| from |
→ |
cons |
(13) |
|
dbl1(0) |
→ |
01 |
(14) |
|
quote(0) |
→ |
01 |
(15) |
|
quote(dbl(X)) |
→ |
dbl1(X) |
(16) |
|
quote(sel(X,Y)) |
→ |
sel1(X,Y) |
(17) |
Property / Task
Prove or disprove termination.Answer / Result
No.Proof (by ttt2 @ termCOMP 2023)
1 Loop
The following loop proves nontermination.
| t0
|
= |
sel(0,cons) |
|
→
|
sel(0,cons) |
|
= |
t1
|
where t1 =
t0σ
and
σ =
{X/sel(0,cons)}