We consider the TRS containing the following rules:
+(0,0) | → | 0 | (1) |
+(s(0),y) | → | s(+(0,y)) | (2) |
+(x,s(y)) | → | s(+(y,x)) | (3) |
s(s(x)) | → | x | (4) |
The underlying signature is as follows:
{+/2, 0/0, s/1}t0 | = | +(f5,s(s(f6))) |
→ | +(f5,f6) | |
= | t1 |
t0 | = | +(f5,s(s(f6))) |
→ | s(+(s(f6),f5)) | |
= | t1 |
Automaton 1
final states:
{3}
transitions:
f5 | → | 5 |
+(5,4) | → | 3 |
f6 | → | 4 |
Automaton 2
final states:
{6}
transitions:
f5 | → | 7 |
s(10) | → | 6 |
s(8) | → | 9 |
+(9,7) | → | 10 |
f6 | → | 8 |