Certification Problem
Input (TPDB TRS_Standard/Rubio_04/prov)
The rewrite relation of the following TRS is considered.
|
ackin(s(X),s(Y)) |
→ |
u21(ackin(s(X),Y),X) |
(1) |
|
u21(ackout(X),Y) |
→ |
u22(ackin(Y,X)) |
(2) |
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(u22) |
= |
0 |
|
status(u22) |
= |
[1] |
|
list-extension(u22) |
= |
Lex |
| prec(ackout) |
= |
0 |
|
status(ackout) |
= |
[1] |
|
list-extension(ackout) |
= |
Lex |
| prec(u21) |
= |
0 |
|
status(u21) |
= |
[2, 1] |
|
list-extension(u21) |
= |
Lex |
| prec(ackin) |
= |
4 |
|
status(ackin) |
= |
[1, 2] |
|
list-extension(ackin) |
= |
Lex |
| prec(s) |
= |
0 |
|
status(s) |
= |
[1] |
|
list-extension(s) |
= |
Lex |
and the following
Max-polynomial interpretation
| [u22(x1)] |
=
|
0 + 1 · x1
|
| [ackout(x1)] |
=
|
max(3, 5 + 1 · x1) |
| [u21(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 6 + 1 · x2) |
| [ackin(x1, x2)] |
=
|
max(0, 4 + 1 · x1, 4 + 1 · x2) |
| [s(x1)] |
=
|
2 + 1 · x1
|
all of the following rules can be deleted.
|
ackin(s(X),s(Y)) |
→ |
u21(ackin(s(X),Y),X) |
(1) |
|
u21(ackout(X),Y) |
→ |
u22(ackin(Y,X)) |
(2) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.