Certification Problem
Input (TPDB TRS_Standard/SK90/2.38)
The rewrite relation of the following TRS is considered.
|
++(nil,y) |
→ |
y |
(1) |
|
++(x,nil) |
→ |
x |
(2) |
|
++(.(x,y),z) |
→ |
.(x,++(y,z)) |
(3) |
|
++(++(x,y),z) |
→ |
++(x,++(y,z)) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(.) |
= |
0 |
|
weight(.) |
= |
0 |
|
|
|
| prec(++) |
= |
1 |
|
weight(++) |
= |
0 |
|
|
|
| prec(nil) |
= |
3 |
|
weight(nil) |
= |
1 |
|
|
|
all of the following rules can be deleted.
|
++(nil,y) |
→ |
y |
(1) |
|
++(x,nil) |
→ |
x |
(2) |
|
++(.(x,y),z) |
→ |
.(x,++(y,z)) |
(3) |
|
++(++(x,y),z) |
→ |
++(x,++(y,z)) |
(4) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.