We consider the TRS containing the following rules:
a(b(a(x))) | → | b(b(b(a(x)))) | (1) |
b(b(b(a(x)))) | → | a(a(a(b(x)))) | (2) |
The underlying signature is as follows:
{a/1, b/1}t0 | = | b(b(b(a(b(a(f3)))))) |
→ | b(b(b(b(b(b(a(f3))))))) | |
= | t1 |
t0 | = | b(b(b(a(b(a(f3)))))) |
→ | a(a(a(b(b(a(f3)))))) | |
= | t1 |
Automaton 1
final states:
{1}
transitions:
b(548) | → | 454 |
b(93) | → | 643 |
b(2) | → | 93 |
b(207) | → | 376 |
b(3) | → | 4 |
b(6) | → | 7 |
b(376) | → | 208 |
b(7) | → | 8 |
b(5) | → | 6 |
b(453) | → | 548 |
b(4) | → | 5 |
b(94) | → | 453 |
b(95) | → | 207 |
b(8) | → | 1 |
a(207) | → | 208 |
a(2) | → | 3 |
a(454) | → | 207 |
a(95) | → | 6 |
a(94) | → | 95 |
a(93) | → | 94 |
a(643) | → | 644 |
a(453) | → | 454 |
a(644) | → | 453 |
a(208) | → | 209 |
a(209) | → | 1 |
f3 | → | 2 |
Automaton 2
final states:
{9}
transitions:
b(11) | → | 12 |
b(12) | → | 13 |
a(14) | → | 15 |
a(15) | → | 9 |
a(10) | → | 11 |
a(13) | → | 14 |
f3 | → | 10 |