Certification Problem
Input (TPDB TRS_Standard/SK90/2.13)
The rewrite relation of the following TRS is considered.
|
double(0) |
→ |
0 |
(1) |
|
double(s(x)) |
→ |
s(s(double(x))) |
(2) |
|
+(x,0) |
→ |
x |
(3) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(4) |
|
+(s(x),y) |
→ |
s(+(x,y)) |
(5) |
|
double(x) |
→ |
+(x,x) |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [double(x1)] |
= |
8 · x1 + 18 |
| [+(x1, x2)] |
= |
2 · x1 + 6 · x2 + 17 |
| [0] |
= |
8 |
| [s(x1)] |
= |
1 · x1 + 17 |
all of the following rules can be deleted.
|
double(0) |
→ |
0 |
(1) |
|
double(s(x)) |
→ |
s(s(double(x))) |
(2) |
|
+(x,0) |
→ |
x |
(3) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(4) |
|
+(s(x),y) |
→ |
s(+(x,y)) |
(5) |
|
double(x) |
→ |
+(x,x) |
(6) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.