Certification Problem
Input (TPDB SRS_Standard/Zantema_06/01)
The rewrite relation of the following TRS is considered.
|
a(b(x1)) |
→ |
b(r(x1)) |
(1) |
|
r(a(x1)) |
→ |
d(r(x1)) |
(2) |
|
r(x1) |
→ |
d(x1) |
(3) |
|
d(a(x1)) |
→ |
a(a(d(x1))) |
(4) |
|
d(x1) |
→ |
a(x1) |
(5) |
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(d) |
= |
2 |
|
status(d) |
= |
[1] |
|
list-extension(d) |
= |
Lex |
| prec(r) |
= |
3 |
|
status(r) |
= |
[1] |
|
list-extension(r) |
= |
Lex |
| prec(a) |
= |
1 |
|
status(a) |
= |
[1] |
|
list-extension(a) |
= |
Lex |
| prec(b) |
= |
0 |
|
status(b) |
= |
[1] |
|
list-extension(b) |
= |
Lex |
and the following
Max-polynomial interpretation
| [d(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [r(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [a(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [b(x1)] |
=
|
4 + 1 · x1
|
all of the following rules can be deleted.
|
a(b(x1)) |
→ |
b(r(x1)) |
(1) |
|
r(a(x1)) |
→ |
d(r(x1)) |
(2) |
|
r(x1) |
→ |
d(x1) |
(3) |
|
d(a(x1)) |
→ |
a(a(d(x1))) |
(4) |
|
d(x1) |
→ |
a(x1) |
(5) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.