Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex25_Luc06_Z)
The rewrite relation of the following TRS is considered.
|
f(f(X)) |
→ |
c(n__f(g(n__f(X)))) |
(1) |
|
c(X) |
→ |
d(activate(X)) |
(2) |
|
h(X) |
→ |
c(n__d(X)) |
(3) |
|
f(X) |
→ |
n__f(X) |
(4) |
|
d(X) |
→ |
n__d(X) |
(5) |
|
activate(n__f(X)) |
→ |
f(X) |
(6) |
|
activate(n__d(X)) |
→ |
d(X) |
(7) |
|
activate(X) |
→ |
X |
(8) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
f(f(X)) |
→ |
n__f(g(n__f(c(X)))) |
(9) |
|
c(X) |
→ |
activate(d(X)) |
(10) |
|
h(X) |
→ |
n__d(c(X)) |
(11) |
|
f(X) |
→ |
n__f(X) |
(4) |
|
d(X) |
→ |
n__d(X) |
(5) |
|
n__f(activate(X)) |
→ |
f(X) |
(12) |
|
n__d(activate(X)) |
→ |
d(X) |
(13) |
|
activate(X) |
→ |
X |
(8) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(n__d) |
= |
7 |
|
weight(n__d) |
= |
0 |
|
|
|
| prec(h) |
= |
2 |
|
weight(h) |
= |
6 |
|
|
|
| prec(d) |
= |
5 |
|
weight(d) |
= |
1 |
|
|
|
| prec(activate) |
= |
4 |
|
weight(activate) |
= |
4 |
|
|
|
| prec(c) |
= |
6 |
|
weight(c) |
= |
5 |
|
|
|
| prec(g) |
= |
0 |
|
weight(g) |
= |
1 |
|
|
|
| prec(n__f) |
= |
1 |
|
weight(n__f) |
= |
1 |
|
|
|
| prec(f) |
= |
3 |
|
weight(f) |
= |
4 |
|
|
|
all of the following rules can be deleted.
|
f(f(X)) |
→ |
n__f(g(n__f(c(X)))) |
(9) |
|
c(X) |
→ |
activate(d(X)) |
(10) |
|
h(X) |
→ |
n__d(c(X)) |
(11) |
|
f(X) |
→ |
n__f(X) |
(4) |
|
d(X) |
→ |
n__d(X) |
(5) |
|
n__f(activate(X)) |
→ |
f(X) |
(12) |
|
n__d(activate(X)) |
→ |
d(X) |
(13) |
|
activate(X) |
→ |
X |
(8) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.