YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
d(a(x0)) |
→ |
b(d(x0)) |
|
b(x0) |
→ |
a(a(a(x0))) |
|
c(d(c(x0))) |
→ |
a(d(x0)) |
|
b(d(d(x0))) |
→ |
c(c(d(d(c(x0))))) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [d(x1)] |
= |
3 ·
x1 + 0 |
| [c(x1)] |
= |
1 ·
x1 + 1 |
| [a(x1)] |
= |
1 ·
x1 + 4 |
| [b(x1)] |
= |
1 ·
x1 + 12 |
the
rules
|
d(a(x0)) |
→ |
b(d(x0)) |
|
b(x0) |
→ |
a(a(a(x0))) |
|
c(d(c(x0))) |
→ |
a(d(x0)) |
remain.
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [d(x1)] |
= |
0 ·
x1 +
-∞
|
| [c(x1)] |
= |
8 ·
x1 +
-∞
|
| [a(x1)] |
= |
0 ·
x1 +
-∞
|
| [b(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
|
d(a(x0)) |
→ |
b(d(x0)) |
|
b(x0) |
→ |
a(a(a(x0))) |
remain.
1.1.1 Bounds
The given TRS is
match-bounded by 1.
This is shown by the following automaton.
-
final states:
{4, 1}
-
transitions:
| 10 |
→ |
1 |
| 3 |
→ |
7 |
| 1 |
→ |
7 |
| 1 |
→ |
3 |
|
a1(7) |
→ |
8 |
|
a1(8) |
→ |
9 |
|
a1(9) |
→ |
10 |
|
f40
|
→ |
2 |
|
d0(2) |
→ |
3 |
|
a0(2) |
→ |
5 |
|
a0(6) |
→ |
4 |
|
a0(5) |
→ |
6 |
|
b0(3) |
→ |
1 |