YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
a(b(c(x0))) |
→ |
c(c(c(b(b(b(a(a(a(x0))))))))) |
|
c(b(x0)) |
→ |
a(a(a(x0))) |
|
a(x0) |
→ |
x0 |
|
b(x0) |
→ |
x0 |
|
c(x0) |
→ |
x0 |
Proof
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
c(b(a(x0))) |
→ |
a(a(a(b(b(b(c(c(c(x0))))))))) |
|
b(c(x0)) |
→ |
a(a(a(x0))) |
|
a(x0) |
→ |
x0 |
|
b(x0) |
→ |
x0 |
|
c(x0) |
→ |
x0 |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
c#(b(a(x0))) |
→ |
c#(x0) |
|
c#(b(a(x0))) |
→ |
c#(c(x0)) |
|
c#(b(a(x0))) |
→ |
c#(c(c(x0))) |
|
c#(b(a(x0))) |
→ |
b#(c(c(c(x0)))) |
|
c#(b(a(x0))) |
→ |
b#(b(c(c(c(x0))))) |
|
c#(b(a(x0))) |
→ |
b#(b(b(c(c(c(x0)))))) |
|
c#(b(a(x0))) |
→ |
a#(b(b(b(c(c(c(x0))))))) |
|
c#(b(a(x0))) |
→ |
a#(a(b(b(b(c(c(c(x0)))))))) |
|
c#(b(a(x0))) |
→ |
a#(a(a(b(b(b(c(c(c(x0))))))))) |
|
b#(c(x0)) |
→ |
a#(x0) |
|
b#(c(x0)) |
→ |
a#(a(x0)) |
|
b#(c(x0)) |
→ |
a#(a(a(x0))) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.