Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-376)
The rewrite relation of the following TRS is considered.
|
a(b(b(b(x1)))) |
→ |
b(a(a(b(x1)))) |
(1) |
|
b(a(a(a(x1)))) |
→ |
b(b(a(a(x1)))) |
(2) |
|
a(b(a(b(x1)))) |
→ |
b(b(a(a(x1)))) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a#(b(b(b(x1)))) |
→ |
b#(a(a(b(x1)))) |
(4) |
|
a#(b(b(b(x1)))) |
→ |
a#(a(b(x1))) |
(5) |
|
a#(b(b(b(x1)))) |
→ |
a#(b(x1)) |
(6) |
|
b#(a(a(a(x1)))) |
→ |
b#(b(a(a(x1)))) |
(7) |
|
b#(a(a(a(x1)))) |
→ |
b#(a(a(x1))) |
(8) |
|
a#(b(a(b(x1)))) |
→ |
b#(b(a(a(x1)))) |
(9) |
|
a#(b(a(b(x1)))) |
→ |
b#(a(a(x1))) |
(10) |
|
a#(b(a(b(x1)))) |
→ |
a#(a(x1)) |
(11) |
|
a#(b(a(b(x1)))) |
→ |
a#(x1) |
(12) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
a#(b(b(b(x1)))) |
→ |
a#(b(x1)) |
(6) |
|
a#(b(b(b(x1)))) |
→ |
a#(a(b(x1))) |
(5) |
|
a#(b(a(b(x1)))) |
→ |
a#(a(x1)) |
(11) |
|
a#(b(a(b(x1)))) |
→ |
a#(x1) |
(12) |
1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [a#(x1)] |
= |
1 · x1
|
| [b(x1)] |
= |
1 + 1 · x1
|
| [a(x1)] |
= |
1 + 1 · x1
|
the
pairs
|
a#(b(b(b(x1)))) |
→ |
a#(b(x1)) |
(6) |
|
a#(b(b(b(x1)))) |
→ |
a#(a(b(x1))) |
(5) |
|
a#(b(a(b(x1)))) |
→ |
a#(a(x1)) |
(11) |
|
a#(b(a(b(x1)))) |
→ |
a#(x1) |
(12) |
could be deleted.
1.1.1.1 P is empty
There are no pairs anymore.
-
The
2nd
component contains the
pair
|
b#(a(a(a(x1)))) |
→ |
b#(a(a(x1))) |
(8) |
1.1.2 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
b#(a(a(a(x1)))) |
→ |
b#(a(a(x1))) |
(8) |
|
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.