Certification Problem
Input (TPDB TRS_Standard/SK90/2.01)
The rewrite relation of the following TRS is considered.
|
i(0) |
→ |
0 |
(1) |
|
+(0,y) |
→ |
y |
(2) |
|
+(x,0) |
→ |
x |
(3) |
|
i(i(x)) |
→ |
x |
(4) |
|
+(i(x),x) |
→ |
0 |
(5) |
|
+(x,i(x)) |
→ |
0 |
(6) |
|
i(+(x,y)) |
→ |
+(i(x),i(y)) |
(7) |
|
+(x,+(y,z)) |
→ |
+(+(x,y),z) |
(8) |
|
+(+(x,i(y)),y) |
→ |
x |
(9) |
|
+(+(x,y),i(y)) |
→ |
x |
(10) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(i) |
= |
1 |
|
stat(i) |
= |
lex
|
| prec(0) |
= |
0 |
|
stat(0) |
= |
lex
|
| prec(+) |
= |
0 |
|
stat(+) |
= |
lex
|
| π(i) |
= |
[1] |
| π(0) |
= |
[] |
| π(+) |
= |
[2,1] |
all of the following rules can be deleted.
|
i(0) |
→ |
0 |
(1) |
|
+(0,y) |
→ |
y |
(2) |
|
+(x,0) |
→ |
x |
(3) |
|
i(i(x)) |
→ |
x |
(4) |
|
+(i(x),x) |
→ |
0 |
(5) |
|
+(x,i(x)) |
→ |
0 |
(6) |
|
i(+(x,y)) |
→ |
+(i(x),i(y)) |
(7) |
|
+(x,+(y,z)) |
→ |
+(+(x,y),z) |
(8) |
|
+(+(x,i(y)),y) |
→ |
x |
(9) |
|
+(+(x,y),i(y)) |
→ |
x |
(10) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.