YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
b(a(a(x0))) |
→ |
a(b(c(x0))) |
|
c(a(x0)) |
→ |
a(c(x0)) |
|
c(b(x0)) |
→ |
b(a(x0)) |
|
L(a(a(x0))) |
→ |
L(a(b(c(x0)))) |
|
c(R(x0)) |
→ |
b(a(R(x0))) |
Proof
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
a(a(b(x0))) |
→ |
c(b(a(x0))) |
|
a(c(x0)) |
→ |
c(a(x0)) |
|
b(c(x0)) |
→ |
a(b(x0)) |
|
a(a(L(x0))) |
→ |
c(b(a(L(x0)))) |
|
R(c(x0)) |
→ |
R(a(b(x0))) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a#(a(b(x0))) |
→ |
a#(x0) |
|
a#(a(b(x0))) |
→ |
b#(a(x0)) |
|
a#(c(x0)) |
→ |
a#(x0) |
|
b#(c(x0)) |
→ |
b#(x0) |
|
b#(c(x0)) |
→ |
a#(b(x0)) |
|
a#(a(L(x0))) |
→ |
b#(a(L(x0))) |
|
R#(c(x0)) |
→ |
b#(x0) |
|
R#(c(x0)) |
→ |
a#(b(x0)) |
|
R#(c(x0)) |
→ |
R#(a(b(x0))) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.