Certification Problem
Input (TPDB TRS_Standard/SK90/2.04)
The rewrite relation of the following TRS is considered.
|
f(+(x,0)) |
→ |
f(x) |
(1) |
|
+(x,+(y,z)) |
→ |
+(+(x,y),z) |
(2) |
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
| [f(x1)] |
= |
2 · x1
|
| [+(x1, x2)] |
= |
1 + 1 · x1 + 2 · x2
|
| [0] |
= |
0 |
all of the following rules can be deleted.
|
f(+(x,0)) |
→ |
f(x) |
(1) |
|
+(x,+(y,z)) |
→ |
+(+(x,y),z) |
(2) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.