Certification Problem
Input (TPDB SRS_Standard/Secret_05_SRS/matchbox1)
The rewrite relation of the following TRS is considered.
|
r(e(x1)) |
→ |
w(r(x1)) |
(1) |
|
i(t(x1)) |
→ |
e(r(x1)) |
(2) |
|
e(w(x1)) |
→ |
r(i(x1)) |
(3) |
|
t(e(x1)) |
→ |
r(e(x1)) |
(4) |
|
w(r(x1)) |
→ |
i(t(x1)) |
(5) |
|
e(r(x1)) |
→ |
e(w(x1)) |
(6) |
|
r(i(t(e(r(x1))))) |
→ |
e(w(r(i(t(e(x1)))))) |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [i(x1)] |
= |
0 · x1 +
-∞ |
| [r(x1)] |
= |
4 · x1 +
-∞ |
| [t(x1)] |
= |
6 · x1 +
-∞ |
| [e(x1)] |
= |
2 · x1 +
-∞ |
| [w(x1)] |
= |
2 · x1 +
-∞ |
all of the following rules can be deleted.
|
t(e(x1)) |
→ |
r(e(x1)) |
(4) |
|
e(r(x1)) |
→ |
e(w(x1)) |
(6) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [i(x1)] |
= |
· x1 +
|
| [r(x1)] |
= |
· x1 +
|
| [t(x1)] |
= |
· x1 +
|
| [e(x1)] |
= |
· x1 +
|
| [w(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
r(i(t(e(r(x1))))) |
→ |
e(w(r(i(t(e(x1)))))) |
(7) |
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(i) |
= |
1 |
|
weight(i) |
= |
2 |
|
|
|
| prec(t) |
= |
4 |
|
weight(t) |
= |
1 |
|
|
|
| prec(w) |
= |
2 |
|
weight(w) |
= |
2 |
|
|
|
| prec(r) |
= |
7 |
|
weight(r) |
= |
1 |
|
|
|
| prec(e) |
= |
0 |
|
weight(e) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
r(e(x1)) |
→ |
w(r(x1)) |
(1) |
|
i(t(x1)) |
→ |
e(r(x1)) |
(2) |
|
e(w(x1)) |
→ |
r(i(x1)) |
(3) |
|
w(r(x1)) |
→ |
i(t(x1)) |
(5) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.