Certification Problem
Input (TPDB SRS_Relative/Zantema_06_relative/rel09)
The relative rewrite relation R/S is considered where R is the following TRS
|
b(q(b(x1))) |
→ |
b(p(b(x1))) |
(1) |
and S is the following TRS.
|
0(p(0(x1))) |
→ |
q(x1) |
(2) |
|
1(q(1(x1))) |
→ |
q(x1) |
(3) |
|
p(x1) |
→ |
1(p(1(0(1(x1))))) |
(4) |
|
0(q(0(x1))) |
→ |
q(x1) |
(5) |
|
1(p(1(x1))) |
→ |
q(x1) |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
matrix interpretations of dimension 2 with strict dimension 1 over the integers
| [b(x1)] |
= |
+ · x1
|
| [q(x1)] |
= |
+ · x1
|
| [p(x1)] |
= |
+ · x1
|
| [0(x1)] |
= |
+ · x1
|
| [1(x1)] |
= |
+ · x1
|
all of the following rules can be deleted.
1.1 Rule Removal
Using the
matrix interpretations of dimension 6 with strict dimension 1 over the integers
| [b(x1)] |
= |
+
|
| 1 |
0 |
0 |
1 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
1 |
|
|
· x1
|
| [q(x1)] |
= |
+
|
| 1 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
1 |
| 0 |
1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
1 |
|
|
· x1
|
| [p(x1)] |
= |
+
|
| 1 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
1 |
1 |
0 |
0 |
| 0 |
1 |
1 |
1 |
0 |
1 |
| 0 |
0 |
0 |
1 |
0 |
0 |
|
|
· x1
|
| [1(x1)] |
= |
+
|
| 1 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
1 |
| 0 |
0 |
0 |
1 |
0 |
0 |
| 0 |
1 |
0 |
0 |
0 |
1 |
|
|
· x1
|
| [0(x1)] |
= |
+
|
| 1 |
0 |
0 |
0 |
0 |
0 |
| 0 |
1 |
0 |
0 |
0 |
1 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
0 |
1 |
0 |
|
|
· x1
|
all of the following rules can be deleted.
|
b(q(b(x1))) |
→ |
b(p(b(x1))) |
(1) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.