Certification Problem
Input (TPDB TRS_Standard/SK90/4.26)
The rewrite relation of the following TRS is considered.
|
rev(nil) |
→ |
nil |
(1) |
|
rev(rev(x)) |
→ |
x |
(2) |
|
rev(++(x,y)) |
→ |
++(rev(y),rev(x)) |
(3) |
|
++(nil,y) |
→ |
y |
(4) |
|
++(x,nil) |
→ |
x |
(5) |
|
++(.(x,y),z) |
→ |
.(x,++(y,z)) |
(6) |
|
++(x,++(y,z)) |
→ |
++(++(x,y),z) |
(7) |
|
make(x) |
→ |
.(x,nil) |
(8) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(rev) |
= |
2 |
|
stat(rev) |
= |
lex
|
| prec(nil) |
= |
3 |
|
stat(nil) |
= |
lex
|
| prec(++) |
= |
1 |
|
stat(++) |
= |
lex
|
| prec(.) |
= |
0 |
|
stat(.) |
= |
lex
|
| prec(make) |
= |
3 |
|
stat(make) |
= |
lex
|
| π(rev) |
= |
[1] |
| π(nil) |
= |
[] |
| π(++) |
= |
[2,1] |
| π(.) |
= |
[2,1] |
| π(make) |
= |
[1] |
all of the following rules can be deleted.
|
rev(nil) |
→ |
nil |
(1) |
|
rev(rev(x)) |
→ |
x |
(2) |
|
rev(++(x,y)) |
→ |
++(rev(y),rev(x)) |
(3) |
|
++(nil,y) |
→ |
y |
(4) |
|
++(x,nil) |
→ |
x |
(5) |
|
++(.(x,y),z) |
→ |
.(x,++(y,z)) |
(6) |
|
++(x,++(y,z)) |
→ |
++(++(x,y),z) |
(7) |
|
make(x) |
→ |
.(x,nil) |
(8) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.