Certification Problem
Input (TPDB TRS_Standard/SK90/4.03)
The rewrite relation of the following TRS is considered.
|
+(x,0) |
→ |
x |
(1) |
|
+(minus(x),x) |
→ |
0 |
(2) |
|
minus(0) |
→ |
0 |
(3) |
|
minus(minus(x)) |
→ |
x |
(4) |
|
minus(+(x,y)) |
→ |
+(minus(y),minus(x)) |
(5) |
|
*(x,1) |
→ |
x |
(6) |
|
*(x,0) |
→ |
0 |
(7) |
|
*(x,+(y,z)) |
→ |
+(*(x,y),*(x,z)) |
(8) |
|
*(x,minus(y)) |
→ |
minus(*(x,y)) |
(9) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(+) |
= |
1 |
|
stat(+) |
= |
mul
|
| prec(0) |
= |
0 |
|
stat(0) |
= |
mul
|
| prec(*) |
= |
2 |
|
stat(*) |
= |
lex
|
| prec(1) |
= |
3 |
|
stat(1) |
= |
mul
|
| π(+) |
= |
[1,2] |
| π(0) |
= |
[] |
| π(minus) |
= |
1 |
| π(*) |
= |
[1,2] |
| π(1) |
= |
[] |
all of the following rules can be deleted.
|
+(x,0) |
→ |
x |
(1) |
|
+(minus(x),x) |
→ |
0 |
(2) |
|
*(x,1) |
→ |
x |
(6) |
|
*(x,0) |
→ |
0 |
(7) |
|
*(x,+(y,z)) |
→ |
+(*(x,y),*(x,z)) |
(8) |
1.1 Rule Removal
Using the
| prec(minus) |
= |
2 |
|
stat(minus) |
= |
mul
|
| prec(0) |
= |
0 |
|
stat(0) |
= |
mul
|
| prec(+) |
= |
1 |
|
stat(+) |
= |
mul
|
| prec(*) |
= |
3 |
|
stat(*) |
= |
lex
|
| π(minus) |
= |
[1] |
| π(0) |
= |
[] |
| π(+) |
= |
[1,2] |
| π(*) |
= |
[1,2] |
all of the following rules can be deleted.
|
minus(0) |
→ |
0 |
(3) |
|
minus(minus(x)) |
→ |
x |
(4) |
|
minus(+(x,y)) |
→ |
+(minus(y),minus(x)) |
(5) |
|
*(x,minus(y)) |
→ |
minus(*(x,y)) |
(9) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.