Certification Problem
Input (TPDB SRS_Standard/Gebhardt_06/11)
The rewrite relation of the following TRS is considered.
|
0(0(0(0(x1)))) |
→ |
0(1(1(1(x1)))) |
(1) |
|
1(0(1(1(x1)))) |
→ |
0(0(0(1(x1)))) |
(2) |
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.
|
0#(0(0(0(x1)))) |
→ |
0#(1(1(1(x1)))) |
(3) |
|
0#(0(0(0(x1)))) |
→ |
1#(1(1(x1))) |
(4) |
|
0#(0(0(0(x1)))) |
→ |
1#(1(x1)) |
(5) |
|
0#(0(0(0(x1)))) |
→ |
1#(x1) |
(6) |
|
1#(0(1(1(x1)))) |
→ |
0#(0(0(1(x1)))) |
(7) |
|
1#(0(1(1(x1)))) |
→ |
0#(0(1(x1))) |
(8) |
|
1#(0(1(1(x1)))) |
→ |
0#(1(x1)) |
(9) |
1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [0#(x1)] |
= |
1 · x1
|
| [0(x1)] |
= |
1 + 1 · x1
|
| [1(x1)] |
= |
1 + 1 · x1
|
| [1#(x1)] |
= |
1 + 1 · x1
|
the
pairs
|
0#(0(0(0(x1)))) |
→ |
1#(1(x1)) |
(5) |
|
0#(0(0(0(x1)))) |
→ |
1#(x1) |
(6) |
|
1#(0(1(1(x1)))) |
→ |
0#(0(0(1(x1)))) |
(7) |
|
1#(0(1(1(x1)))) |
→ |
0#(0(1(x1))) |
(8) |
|
1#(0(1(1(x1)))) |
→ |
0#(1(x1)) |
(9) |
could be deleted.
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.