Certification Problem
Input (TPDB TRS_Standard/SK90/2.10)
The rewrite relation of the following TRS is considered.
|
minus(0) |
→ |
0 |
(1) |
|
+(x,0) |
→ |
x |
(2) |
|
+(0,y) |
→ |
y |
(3) |
|
+(minus(1),1) |
→ |
0 |
(4) |
|
minus(minus(x)) |
→ |
x |
(5) |
|
+(x,minus(y)) |
→ |
minus(+(minus(x),y)) |
(6) |
|
+(x,+(y,z)) |
→ |
+(+(x,y),z) |
(7) |
|
+(minus(+(x,1)),1) |
→ |
minus(x) |
(8) |
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
| [minus(x1)] |
= |
2 + 1 · x1
|
| [0] |
= |
0 |
| [+(x1, x2)] |
= |
1 + 1 · x1 + 2 · x2
|
| [1] |
= |
0 |
all of the following rules can be deleted.
|
minus(0) |
→ |
0 |
(1) |
|
+(x,0) |
→ |
x |
(2) |
|
+(0,y) |
→ |
y |
(3) |
|
+(minus(1),1) |
→ |
0 |
(4) |
|
minus(minus(x)) |
→ |
x |
(5) |
|
+(x,+(y,z)) |
→ |
+(+(x,y),z) |
(7) |
|
+(minus(+(x,1)),1) |
→ |
minus(x) |
(8) |
1.1 Rule Removal
Using the
| prec(+) |
= |
1 |
|
stat(+) |
= |
lex
|
| prec(minus) |
= |
0 |
|
stat(minus) |
= |
lex
|
| π(+) |
= |
[2,1] |
| π(minus) |
= |
[1] |
all of the following rules can be deleted.
|
+(x,minus(y)) |
→ |
minus(+(minus(x),y)) |
(6) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.