Certification Problem
Input (TPDB SRS_Relative/Zantema_06_relative/rel08)
The relative rewrite relation R/S is considered where R is the following TRS
|
a(x1) |
→ |
x1 |
(1) |
|
b(x1) |
→ |
x1 |
(2) |
|
a(c(x1)) |
→ |
b(b(c(a(x1)))) |
(3) |
|
d(b(b(b(x1)))) |
→ |
a(d(a(x1))) |
(4) |
and S is the following TRS.
|
a(b(x1)) |
→ |
b(a(x1)) |
(5) |
|
b(a(x1)) |
→ |
a(b(x1)) |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by matchbox @ termCOMP 2023)
1 Rule Removal
Using the
matrix interpretations of dimension 3 with strict dimension 1 over the naturals
| [d(x1)] |
= |
· x1 +
|
| [c(x1)] |
= |
· x1 +
|
| [b(x1)] |
= |
· x1 +
|
| [a(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
b(x1) |
→ |
x1 |
(2) |
|
d(b(b(b(x1)))) |
→ |
a(d(a(x1))) |
(4) |
1.1 Rule Removal
Using the
matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
| [c(x1)] |
= |
x1 +
|
| [b(x1)] |
= |
x1 +
|
| [a(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
1.1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
c(a(x1)) |
→ |
a(c(b(b(x1)))) |
(7) |
|
b(a(x1)) |
→ |
a(b(x1)) |
(6) |
|
a(b(x1)) |
→ |
b(a(x1)) |
(5) |
1.1.1.1 Rule Removal
Using the
matrix interpretations of dimension 1 with strict dimension 1 over the naturals
| [c(x1)] |
= |
· x1 +
|
| [b(x1)] |
= |
· x1 +
|
| [a(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
c(a(x1)) |
→ |
a(c(b(b(x1)))) |
(7) |
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.