Certification Problem
Input (TPDB TRS_Standard/Der95/07)
The rewrite relation of the following TRS is considered.
|
w(r(x)) |
→ |
r(w(x)) |
(1) |
|
b(r(x)) |
→ |
r(b(x)) |
(2) |
|
b(w(x)) |
→ |
w(b(x)) |
(3) |
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
|
r(w(x)) |
→ |
w(r(x)) |
(4) |
|
r(b(x)) |
→ |
b(r(x)) |
(5) |
|
w(b(x)) |
→ |
b(w(x)) |
(6) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(b) |
= |
0 |
|
weight(b) |
= |
2 |
|
|
|
| prec(w) |
= |
2 |
|
weight(w) |
= |
2 |
|
|
|
| prec(r) |
= |
3 |
|
weight(r) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
r(w(x)) |
→ |
w(r(x)) |
(4) |
|
r(b(x)) |
→ |
b(r(x)) |
(5) |
|
w(b(x)) |
→ |
b(w(x)) |
(6) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.