We consider the TRS containing the following rules:
a(a(b(b(x)))) | → | b(b(b(a(a(a(x)))))) | (1) |
c(x) | → | c(a(x)) | (2) |
c(x) | → | c(b(x)) | (3) |
The underlying signature is as follows:
{a/1, b/1, c/1}t0 | = | c(f15) |
→ | c(b(f15)) | |
→ | c(a(b(f15))) | |
= | t2 |
t0 | = | c(f15) |
→ | c(a(f15)) | |
= | t1 |
Automaton 1
final states:
{15}
transitions:
f15 | → | 16 |
c(18) | → | 15 |
a(18) | → | 18 |
a(17) | → | 18 |
b(18) | → | 18 |
b(16) | → | 17 |
Automaton 2
final states:
{4}
transitions:
f15 | → | 5 |
c(6) | → | 4 |
a(6) | → | 6 |
a(5) | → | 6 |
b(6) | → | 6 |