Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex2_Luc03b_L)
The rewrite relation of the following TRS is considered.
|
fst(0,Z) |
→ |
nil |
(1) |
|
fst(s,cons(Y)) |
→ |
cons(Y) |
(2) |
|
from(X) |
→ |
cons(X) |
(3) |
|
add(0,X) |
→ |
X |
(4) |
|
add(s,Y) |
→ |
s |
(5) |
|
len(nil) |
→ |
0 |
(6) |
|
len(cons(X)) |
→ |
s |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [fst(x1, x2)] |
= |
1 · x1 + 2 · x2
|
| [0] |
= |
1 |
| [nil] |
= |
0 |
| [s] |
= |
1 |
| [cons(x1)] |
= |
1 · x1
|
| [from(x1)] |
= |
1 + 2 · x1
|
| [add(x1, x2)] |
= |
2 · x1 + 1 · x2
|
| [len(x1)] |
= |
2 + 1 · x1
|
all of the following rules can be deleted.
|
fst(0,Z) |
→ |
nil |
(1) |
|
fst(s,cons(Y)) |
→ |
cons(Y) |
(2) |
|
from(X) |
→ |
cons(X) |
(3) |
|
add(0,X) |
→ |
X |
(4) |
|
add(s,Y) |
→ |
s |
(5) |
|
len(nil) |
→ |
0 |
(6) |
|
len(cons(X)) |
→ |
s |
(7) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.