YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
a(b(a(x0))) |
→ |
b(b(b(a(x0)))) |
|
b(b(b(a(x0)))) |
→ |
a(a(a(b(x0)))) |
Proof
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
a(b(a(x0))) |
→ |
a(b(b(b(x0)))) |
|
a(b(b(b(x0)))) |
→ |
b(a(a(a(x0)))) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a#(b(a(x0))) |
→ |
a#(b(b(b(x0)))) |
|
a#(b(b(b(x0)))) |
→ |
a#(x0) |
|
a#(b(b(b(x0)))) |
→ |
a#(a(x0)) |
|
a#(b(b(b(x0)))) |
→ |
a#(a(a(x0))) |
1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1
over the naturals
| [b(x1)] |
= |
| 0 |
1 |
1 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
1 |
| 1 |
0 |
0 |
0 |
·
x1 +
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
|
| [a(x1)] |
= |
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
·
x1 +
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
| [a#(x1)] |
= |
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
·
x1 +
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
the
pair
|
a#(b(a(x0))) |
→ |
a#(b(b(b(x0)))) |
remains.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.