Certification Problem
Input (TPDB SRS_Standard/Waldmann_06_SRS/jw4)
The rewrite relation of the following TRS is considered.
|
b(a(b(b(x1)))) |
→ |
b(b(b(a(b(x1))))) |
(1) |
|
b(a(a(b(b(x1))))) |
→ |
b(a(b(b(a(a(b(x1))))))) |
(2) |
|
b(a(a(a(b(b(x1)))))) |
→ |
b(a(a(b(b(a(a(a(b(x1))))))))) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(a(b(b(x1)))) |
→ |
b#(a(b(x1))) |
(4) |
|
b#(a(b(b(x1)))) |
→ |
b#(b(a(b(x1)))) |
(5) |
|
b#(a(b(b(x1)))) |
→ |
b#(b(b(a(b(x1))))) |
(6) |
|
b#(a(a(b(b(x1))))) |
→ |
b#(a(a(b(x1)))) |
(7) |
|
b#(a(a(b(b(x1))))) |
→ |
b#(b(a(a(b(x1))))) |
(8) |
|
b#(a(a(b(b(x1))))) |
→ |
b#(a(b(b(a(a(b(x1))))))) |
(9) |
|
b#(a(a(a(b(b(x1)))))) |
→ |
b#(a(a(a(b(x1))))) |
(10) |
|
b#(a(a(a(b(b(x1)))))) |
→ |
b#(b(a(a(a(b(x1)))))) |
(11) |
|
b#(a(a(a(b(b(x1)))))) |
→ |
b#(a(a(b(b(a(a(a(b(x1))))))))) |
(12) |
1.1 Dependency Graph Processor
The dependency pairs are split into 3
components.
-
The
1st
component contains the
pair
|
b#(a(a(a(b(b(x1)))))) |
→ |
b#(a(a(a(b(x1))))) |
(10) |
1.1.1 Subterm Criterion Processor
We use the projection to multisets
| π(b#)
|
= |
{
1
}
|
| π(a)
|
= |
{
1
}
|
to remove the pairs:
|
b#(a(a(a(b(b(x1)))))) |
→ |
b#(a(a(a(b(x1))))) |
(10) |
1.1.1.1 P is empty
There are no pairs anymore.
-
The
2nd
component contains the
pair
|
b#(a(a(b(b(x1))))) |
→ |
b#(a(a(b(x1)))) |
(7) |
1.1.2 Subterm Criterion Processor
We use the projection to multisets
| π(b#)
|
= |
{
1
}
|
| π(a)
|
= |
{
1
}
|
to remove the pairs:
|
b#(a(a(b(b(x1))))) |
→ |
b#(a(a(b(x1)))) |
(7) |
1.1.2.1 P is empty
There are no pairs anymore.
-
The
3rd
component contains the
pair
|
b#(a(b(b(x1)))) |
→ |
b#(a(b(x1))) |
(4) |
1.1.3 Subterm Criterion Processor
We use the projection to multisets
| π(b#)
|
= |
{
1
}
|
| π(a)
|
= |
{
1
}
|
to remove the pairs:
|
b#(a(b(b(x1)))) |
→ |
b#(a(b(x1))) |
(4) |
1.1.3.1 P is empty
There are no pairs anymore.