Certification Problem
Input (TPDB TRS_Standard/SK90/4.10)
The rewrite relation of the following TRS is considered.
|
*(x,*(y,z)) |
→ |
*(otimes(x,y),z) |
(1) |
|
*(1,y) |
→ |
y |
(2) |
|
*(+(x,y),z) |
→ |
oplus(*(x,z),*(y,z)) |
(3) |
|
*(x,oplus(y,z)) |
→ |
oplus(*(x,y),*(x,z)) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(*) |
= |
2 |
|
stat(*) |
= |
lex
|
| prec(otimes) |
= |
0 |
|
stat(otimes) |
= |
mul
|
| prec(1) |
= |
3 |
|
stat(1) |
= |
mul
|
| prec(+) |
= |
4 |
|
stat(+) |
= |
lex
|
| prec(oplus) |
= |
1 |
|
stat(oplus) |
= |
lex
|
| π(*) |
= |
[2,1] |
| π(otimes) |
= |
[1,2] |
| π(1) |
= |
[] |
| π(+) |
= |
[1,2] |
| π(oplus) |
= |
[2,1] |
all of the following rules can be deleted.
|
*(x,*(y,z)) |
→ |
*(otimes(x,y),z) |
(1) |
|
*(1,y) |
→ |
y |
(2) |
|
*(+(x,y),z) |
→ |
oplus(*(x,z),*(y,z)) |
(3) |
|
*(x,oplus(y,z)) |
→ |
oplus(*(x,y),*(x,z)) |
(4) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.