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 AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(ackin) |
= |
1 |
|
stat(ackin) |
= |
lex
|
| prec(s) |
= |
0 |
|
stat(s) |
= |
lex
|
| prec(u21) |
= |
1 |
|
stat(u21) |
= |
lex
|
| π(ackin) |
= |
[1,2] |
| π(s) |
= |
[1] |
| π(u21) |
= |
[2,1] |
| π(ackout) |
= |
1 |
| π(u22) |
= |
1 |
all of the following rules can be deleted.
|
ackin(s(X),s(Y)) |
→ |
u21(ackin(s(X),Y),X) |
(1) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(ackout) |
= |
0 |
|
weight(ackout) |
= |
1 |
|
|
|
| prec(u22) |
= |
1 |
|
weight(u22) |
= |
1 |
|
|
|
| prec(u21) |
= |
2 |
|
weight(u21) |
= |
0 |
|
|
|
| prec(ackin) |
= |
3 |
|
weight(ackin) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
u21(ackout(X),Y) |
→ |
u22(ackin(Y,X)) |
(2) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.