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