YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
a(b(x0)) |
→ |
b(c(a(x0))) |
|
b(c(x0)) |
→ |
c(b(b(x0))) |
|
a(c(x0)) |
→ |
c(a(b(x0))) |
|
a(a(x0)) |
→ |
a(d(d(d(x0)))) |
|
d(a(x0)) |
→ |
d(d(c(x0))) |
|
a(d(d(c(x0)))) |
→ |
a(a(a(d(x0)))) |
|
e(e(f(f(x0)))) |
→ |
f(f(f(e(e(x0))))) |
|
e(x0) |
→ |
a(x0) |
|
b(d(x0)) |
→ |
d(d(x0)) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [f(x1)] |
= |
0 ·
x1 +
-∞
|
| [a(x1)] |
= |
0 ·
x1 +
-∞
|
| [d(x1)] |
= |
0 ·
x1 +
-∞
|
| [e(x1)] |
= |
1 ·
x1 +
-∞
|
| [b(x1)] |
= |
0 ·
x1 +
-∞
|
| [c(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
|
a(b(x0)) |
→ |
b(c(a(x0))) |
|
b(c(x0)) |
→ |
c(b(b(x0))) |
|
a(c(x0)) |
→ |
c(a(b(x0))) |
|
a(a(x0)) |
→ |
a(d(d(d(x0)))) |
|
d(a(x0)) |
→ |
d(d(c(x0))) |
|
a(d(d(c(x0)))) |
→ |
a(a(a(d(x0)))) |
|
e(e(f(f(x0)))) |
→ |
f(f(f(e(e(x0))))) |
|
b(d(x0)) |
→ |
d(d(x0)) |
remain.
1.1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the naturals
| [f(x1)] |
= |
·
x1 +
|
| [a(x1)] |
= |
·
x1 +
|
| [d(x1)] |
= |
·
x1 +
|
| [e(x1)] |
= |
·
x1 +
|
| [b(x1)] |
= |
·
x1 +
|
| [c(x1)] |
= |
·
x1 +
|
the
rules
|
a(b(x0)) |
→ |
b(c(a(x0))) |
|
b(c(x0)) |
→ |
c(b(b(x0))) |
|
a(c(x0)) |
→ |
c(a(b(x0))) |
|
a(a(x0)) |
→ |
a(d(d(d(x0)))) |
|
d(a(x0)) |
→ |
d(d(c(x0))) |
|
a(d(d(c(x0)))) |
→ |
a(a(a(d(x0)))) |
|
b(d(x0)) |
→ |
d(d(x0)) |
remain.
1.1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
b(a(x0)) |
→ |
a(c(b(x0))) |
|
c(b(x0)) |
→ |
b(b(c(x0))) |
|
c(a(x0)) |
→ |
b(a(c(x0))) |
|
a(a(x0)) |
→ |
d(d(d(a(x0)))) |
|
a(d(x0)) |
→ |
c(d(d(x0))) |
|
c(d(d(a(x0)))) |
→ |
d(a(a(a(x0)))) |
|
d(b(x0)) |
→ |
d(d(x0)) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [a(x1)] |
= |
·
x1 +
|
| [d(x1)] |
= |
·
x1 +
|
| [b(x1)] |
= |
·
x1 +
|
| [c(x1)] |
= |
·
x1 +
|
the
rules
|
b(a(x0)) |
→ |
a(c(b(x0))) |
|
c(b(x0)) |
→ |
b(b(c(x0))) |
|
c(a(x0)) |
→ |
b(a(c(x0))) |
|
a(a(x0)) |
→ |
d(d(d(a(x0)))) |
|
a(d(x0)) |
→ |
c(d(d(x0))) |
|
c(d(d(a(x0)))) |
→ |
d(a(a(a(x0)))) |
remain.
1.1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(a(x0)) |
→ |
b#(x0) |
|
b#(a(x0)) |
→ |
c#(b(x0)) |
|
b#(a(x0)) |
→ |
a#(c(b(x0))) |
|
c#(b(x0)) |
→ |
c#(x0) |
|
c#(b(x0)) |
→ |
b#(c(x0)) |
|
c#(b(x0)) |
→ |
b#(b(c(x0))) |
|
c#(a(x0)) |
→ |
c#(x0) |
|
c#(a(x0)) |
→ |
a#(c(x0)) |
|
c#(a(x0)) |
→ |
b#(a(c(x0))) |
|
a#(d(x0)) |
→ |
c#(d(d(x0))) |
|
c#(d(d(a(x0)))) |
→ |
a#(a(x0)) |
|
c#(d(d(a(x0)))) |
→ |
a#(a(a(x0))) |
1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.