YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
a(x0) |
→ |
b(b(x0)) |
|
a(b(b(x0))) |
→ |
b(b(c(c(c(a(x0)))))) |
|
b(b(x0)) |
→ |
c(c(c(x0))) |
|
c(c(c(b(b(x0))))) |
→ |
a(x0) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [b(x1)] |
= |
1 ·
x1 +
-∞
|
| [a(x1)] |
= |
2 ·
x1 +
-∞
|
| [c(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
|
a(x0) |
→ |
b(b(x0)) |
|
a(b(b(x0))) |
→ |
b(b(c(c(c(a(x0)))))) |
|
c(c(c(b(b(x0))))) |
→ |
a(x0) |
remain.
1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
a(x0) |
→ |
b(b(x0)) |
|
b(b(a(x0))) |
→ |
a(c(c(c(b(b(x0)))))) |
|
b(b(c(c(c(x0))))) |
→ |
a(x0) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b(x1)] |
= |
2 ·
x1 + 0 |
| [a(x1)] |
= |
4 ·
x1 + 4 |
| [c(x1)] |
= |
1 ·
x1 + 1 |
the
rule
|
b(b(a(x0))) |
→ |
a(c(c(c(b(b(x0)))))) |
remains.
1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight function
| prec(c) |
= |
3 |
|
weight(c) |
= |
0 |
|
|
|
| prec(b) |
= |
2 |
|
weight(b) |
= |
1 |
|
|
|
| prec(a) |
= |
0 |
|
weight(a) |
= |
1 |
|
|
|
all rules could be removed.
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.