YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
a(x0) |
→ |
x0 |
|
a(x0) |
→ |
b(x0) |
|
a(b(c(x0))) |
→ |
c(c(a(b(a(x0))))) |
|
c(x0) |
→ |
x0 |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the arctic semiring over the integers
| [b(x1)] |
= |
·
x1 +
|
-∞
|
-∞
|
-∞
|
|
-∞
|
-∞
|
-∞
|
|
-∞
|
-∞
|
-∞
|
|
| [a(x1)] |
= |
·
x1 +
|
-∞
|
-∞
|
-∞
|
|
-∞
|
-∞
|
-∞
|
|
-∞
|
-∞
|
-∞
|
|
| [c(x1)] |
= |
·
x1 +
|
-∞
|
-∞
|
-∞
|
|
-∞
|
-∞
|
-∞
|
|
-∞
|
-∞
|
-∞
|
|
the
rules
|
a(x0) |
→ |
x0 |
|
a(b(c(x0))) |
→ |
c(c(a(b(a(x0))))) |
|
c(x0) |
→ |
x0 |
remain.
1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
a(x0) |
→ |
x0 |
|
c(b(a(x0))) |
→ |
a(b(a(c(c(x0))))) |
|
c(x0) |
→ |
x0 |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the naturals
| [b(x1)] |
= |
·
x1 +
|
| [a(x1)] |
= |
·
x1 +
|
| [c(x1)] |
= |
·
x1 +
|
the
rules
remain.
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the arctic semiring over the integers
| [a(x1)] |
= |
·
x1 +
|
-∞
|
-∞
|
-∞
|
|
-∞
|
-∞
|
-∞
|
|
-∞
|
-∞
|
-∞
|
|
| [c(x1)] |
= |
·
x1 +
|
-∞
|
-∞
|
-∞
|
|
-∞
|
-∞
|
-∞
|
|
-∞
|
-∞
|
-∞
|
|
the
rule
remains.
1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the arctic semiring over the integers
| [a(x1)] |
= |
·
x1 +
|
-∞
|
-∞
|
-∞
|
|
-∞
|
-∞
|
-∞
|
|
-∞
|
-∞
|
-∞
|
|
all rules could be removed.
1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.