We consider the TRS containing the following rules:
+(0,y) | → | y | (1) |
+(s(x),y) | → | s(+(x,y)) | (2) |
+(p(x),y) | → | +(x,p(y)) | (3) |
p(s(x)) | → | s(p(x)) | (4) |
s(p(x)) | → | p(s(x)) | (5) |
p(s(0)) | → | 0 | (6) |
The underlying signature is as follows:
{+/2, 0/0, s/1, p/1}t0 | = | +(p(s(0)),f5) |
→ | +(0,f5) | |
= | t1 |
t0 | = | +(p(s(0)),f5) |
→ | +(s(0),p(f5)) | |
= | t1 |
Automaton 1
final states:
{11}
transitions:
12 | → | 11 |
f5 | → | 12 |
+(13,12) | → | 11 |
0 | → | 13 |
Automaton 2
final states:
{14}
transitions:
16 | → | 107 |
f5 | → | 15 |
s(17) | → | 18 |
s(107) | → | 14 |
s(15) | → | 277 |
+(18,16) | → | 14 |
+(17,16) | → | 107 |
0 | → | 17 |
p(15) | → | 16 |
p(277) | → | 14 |