Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z071)
The rewrite relation of the following TRS is considered.
|
R(x1) |
→ |
r(x1) |
(1) |
|
r(p(x1)) |
→ |
p(p(r(P(x1)))) |
(2) |
|
r(r(x1)) |
→ |
x1 |
(3) |
|
r(P(P(x1))) |
→ |
P(P(r(x1))) |
(4) |
|
p(P(x1)) |
→ |
x1 |
(5) |
|
P(p(x1)) |
→ |
x1 |
(6) |
|
r(R(x1)) |
→ |
x1 |
(7) |
|
R(r(x1)) |
→ |
x1 |
(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
|
R(x1) |
→ |
r(x1) |
(1) |
|
p(r(x1)) |
→ |
P(r(p(p(x1)))) |
(9) |
|
r(r(x1)) |
→ |
x1 |
(3) |
|
P(P(r(x1))) |
→ |
r(P(P(x1))) |
(10) |
|
P(p(x1)) |
→ |
x1 |
(6) |
|
p(P(x1)) |
→ |
x1 |
(5) |
|
R(r(x1)) |
→ |
x1 |
(8) |
|
r(R(x1)) |
→ |
x1 |
(7) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [r(x1)] |
= |
0 · x1 +
-∞ |
| [P(x1)] |
= |
0 · x1 +
-∞ |
| [R(x1)] |
= |
2 · x1 +
-∞ |
| [p(x1)] |
= |
0 · x1 +
-∞ |
all of the following rules can be deleted.
|
R(x1) |
→ |
r(x1) |
(1) |
|
R(r(x1)) |
→ |
x1 |
(8) |
|
r(R(x1)) |
→ |
x1 |
(7) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [r(x1)] |
= |
· x1 +
|
| [P(x1)] |
= |
· x1 +
|
| [p(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [r(x1)] |
= |
· x1 +
|
| [P(x1)] |
= |
· x1 +
|
| [p(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
p(r(x1)) |
→ |
P(r(p(p(x1)))) |
(9) |
1.1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(P) |
= |
3 |
|
weight(P) |
= |
2 |
|
|
|
| prec(p) |
= |
0 |
|
weight(p) |
= |
2 |
|
|
|
| prec(r) |
= |
2 |
|
weight(r) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
P(P(r(x1))) |
→ |
r(P(P(x1))) |
(10) |
|
P(p(x1)) |
→ |
x1 |
(6) |
|
p(P(x1)) |
→ |
x1 |
(5) |
1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.