Certification Problem
Input (TPDB TRS_Standard/Der95/08)
The rewrite relation of the following TRS is considered.
|
D(t) |
→ |
1 |
(1) |
|
D(constant) |
→ |
0 |
(2) |
|
D(+(x,y)) |
→ |
+(D(x),D(y)) |
(3) |
|
D(*(x,y)) |
→ |
+(*(y,D(x)),*(x,D(y))) |
(4) |
|
D(-(x,y)) |
→ |
-(D(x),D(y)) |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(D) |
= |
4 |
|
stat(D) |
= |
mul
|
| prec(t) |
= |
5 |
|
stat(t) |
= |
mul
|
| prec(1) |
= |
0 |
|
stat(1) |
= |
mul
|
| prec(constant) |
= |
6 |
|
stat(constant) |
= |
mul
|
| prec(0) |
= |
1 |
|
stat(0) |
= |
mul
|
| prec(+) |
= |
2 |
|
stat(+) |
= |
mul
|
| prec(*) |
= |
2 |
|
stat(*) |
= |
mul
|
| prec(-) |
= |
3 |
|
stat(-) |
= |
mul
|
| π(D) |
= |
[1] |
| π(t) |
= |
[] |
| π(1) |
= |
[] |
| π(constant) |
= |
[] |
| π(0) |
= |
[] |
| π(+) |
= |
[1,2] |
| π(*) |
= |
[1,2] |
| π(-) |
= |
[1,2] |
all of the following rules can be deleted.
|
D(t) |
→ |
1 |
(1) |
|
D(constant) |
→ |
0 |
(2) |
|
D(+(x,y)) |
→ |
+(D(x),D(y)) |
(3) |
|
D(*(x,y)) |
→ |
+(*(y,D(x)),*(x,D(y))) |
(4) |
|
D(-(x,y)) |
→ |
-(D(x),D(y)) |
(5) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.