YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
a(a(x0)) |
→ |
b(c(c(c(x0)))) |
b(c(x0)) |
→ |
d(d(d(d(x0)))) |
a(x0) |
→ |
d(c(d(x0))) |
b(b(x0)) |
→ |
c(c(c(x0))) |
c(c(x0)) |
→ |
d(d(d(x0))) |
c(d(d(x0))) |
→ |
a(x0) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
[c(x1)] |
= |
3 ·
x1 +
-∞
|
[d(x1)] |
= |
2 ·
x1 +
-∞
|
[a(x1)] |
= |
7 ·
x1 +
-∞
|
[b(x1)] |
= |
5 ·
x1 +
-∞
|
the
rules
a(a(x0)) |
→ |
b(c(c(c(x0)))) |
b(c(x0)) |
→ |
d(d(d(d(x0)))) |
a(x0) |
→ |
d(c(d(x0))) |
c(c(x0)) |
→ |
d(d(d(x0))) |
c(d(d(x0))) |
→ |
a(x0) |
remain.
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
[c(x1)] |
= |
4 ·
x1 +
-∞
|
[d(x1)] |
= |
2 ·
x1 +
-∞
|
[a(x1)] |
= |
8 ·
x1 +
-∞
|
[b(x1)] |
= |
4 ·
x1 +
-∞
|
the
rules
a(a(x0)) |
→ |
b(c(c(c(x0)))) |
b(c(x0)) |
→ |
d(d(d(d(x0)))) |
a(x0) |
→ |
d(c(d(x0))) |
c(d(d(x0))) |
→ |
a(x0) |
remain.
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
a#(a(x0)) |
→ |
c#(x0) |
a#(a(x0)) |
→ |
c#(c(x0)) |
a#(a(x0)) |
→ |
c#(c(c(x0))) |
a#(a(x0)) |
→ |
b#(c(c(c(x0)))) |
a#(x0) |
→ |
c#(d(x0)) |
c#(d(d(x0))) |
→ |
a#(x0) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.