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(b(c(x0))) |
b(a(x0)) |
→ |
c(x0) |
c(b(x0)) |
→ |
a(a(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
[c(x1)] |
= |
·
x1 +
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
|
[a(x1)] |
= |
·
x1 +
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
|
[b(x1)] |
= |
·
x1 +
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
-∞
|
|
the
rules
a(a(x0)) |
→ |
b(b(c(x0))) |
c(b(x0)) |
→ |
a(a(x0)) |
remain.
1.1 Bounds
The given TRS is
match-bounded by 1.
This is shown by the following automaton.
-
final states:
{5, 1}
-
transitions:
5 |
→ |
3 |
5 |
→ |
12 |
1 |
→ |
6 |
2 |
→ |
11 |
14 |
→ |
5 |
a0(2) |
→ |
6 |
a0(6) |
→ |
5 |
b0(4) |
→ |
1 |
b0(3) |
→ |
4 |
c1(11) |
→ |
12 |
b1(12) |
→ |
13 |
b1(13) |
→ |
14 |
f30
|
→ |
2 |
c0(2) |
→ |
3 |