Certification Problem
Input (TPDB TRS_Standard/AG01/#3.33)
The rewrite relation of the following TRS is considered.
|
p(f(f(x))) |
→ |
q(f(g(x))) |
(1) |
|
p(g(g(x))) |
→ |
q(g(f(x))) |
(2) |
|
q(f(f(x))) |
→ |
p(f(g(x))) |
(3) |
|
q(g(g(x))) |
→ |
p(g(f(x))) |
(4) |
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(p(x))) |
→ |
g(f(q(x))) |
(5) |
|
g(g(p(x))) |
→ |
f(g(q(x))) |
(6) |
|
f(f(q(x))) |
→ |
g(f(p(x))) |
(7) |
|
g(g(q(x))) |
→ |
f(g(p(x))) |
(8) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [p(x1)] |
= |
· x1 +
|
| [q(x1)] |
= |
· x1 +
|
| [f(x1)] |
= |
· x1 +
|
| [g(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
g(g(q(x))) |
→ |
f(g(p(x))) |
(8) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [p(x1)] |
= |
· x1 +
|
| [q(x1)] |
= |
· x1 +
|
| [f(x1)] |
= |
· x1 +
|
| [g(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
f(f(q(x))) |
→ |
g(f(p(x))) |
(7) |
1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(q) |
= |
2 |
|
weight(q) |
= |
2 |
|
|
|
| prec(g) |
= |
0 |
|
weight(g) |
= |
5 |
|
|
|
| prec(p) |
= |
3 |
|
weight(p) |
= |
5 |
|
|
|
| prec(f) |
= |
1 |
|
weight(f) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
f(f(p(x))) |
→ |
g(f(q(x))) |
(5) |
|
g(g(p(x))) |
→ |
f(g(q(x))) |
(6) |
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.