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 ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
| prec(oplus) |
= |
0 |
|
status(oplus) |
= |
[1, 2] |
|
list-extension(oplus) |
= |
Lex |
| prec(+) |
= |
0 |
|
status(+) |
= |
[2, 1] |
|
list-extension(+) |
= |
Lex |
| prec(1) |
= |
0 |
|
status(1) |
= |
[] |
|
list-extension(1) |
= |
Lex |
| prec(otimes) |
= |
0 |
|
status(otimes) |
= |
[2, 1] |
|
list-extension(otimes) |
= |
Lex |
| prec(*) |
= |
5 |
|
status(*) |
= |
[2, 1] |
|
list-extension(*) |
= |
Lex |
and the following
Max-polynomial interpretation
| [oplus(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 1 + 1 · x2) |
| [+(x1, x2)] |
=
|
4 + 1 · x1 + 1 · x2
|
| [1] |
=
|
max(4) |
| [otimes(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 0 + 1 · x2) |
| [*(x1, x2)] |
=
|
0 + 1 · x1 + 1 · x2
|
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.