YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
a(d(x0)) |
→ |
d(b(x0)) |
|
a(x0) |
→ |
b(b(b(x0))) |
|
b(d(b(x0))) |
→ |
a(c(x0)) |
|
c(x0) |
→ |
d(x0) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [a(x1)] |
= |
1 ·
x1 + 13 |
| [c(x1)] |
= |
13 ·
x1 + 5 |
| [d(x1)] |
= |
13 ·
x1 + 4 |
| [b(x1)] |
= |
1 ·
x1 + 1 |
the
rules
|
a(d(x0)) |
→ |
d(b(x0)) |
|
b(d(b(x0))) |
→ |
a(c(x0)) |
remain.
1.1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the arctic semiring over the integers
| [a(x1)] |
= |
·
x1 +
|
| [c(x1)] |
= |
·
x1 +
|
| [d(x1)] |
= |
·
x1 +
|
| [b(x1)] |
= |
·
x1 +
|
the
rule
remains.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the arctic semiring over the integers
| [a(x1)] |
= |
·
x1 +
|
| [d(x1)] |
= |
·
x1 +
|
| [b(x1)] |
= |
·
x1 +
|
all rules could be removed.
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.