We consider the TRS containing the following rules:
+(0,y) | → | y | (1) |
+(s(x),y) | → | s(+(x,y)) | (2) |
+(p(x),y) | → | p(+(y,x)) | (3) |
p(s(x)) | → | s(p(x)) | (4) |
s(p(x)) | → | x | (5) |
The underlying signature is as follows:
{+/2, 0/0, s/1, p/1}t0 | = | +(s(p(f7)),f5) |
→ | +(f7,f5) | |
= | t1 |
t0 | = | +(s(p(f7)),f5) |
→ | s(+(p(f7),f5)) | |
= | t1 |
Automaton 1
final states:
{7}
transitions:
f7 | → | 9 |
f5 | → | 8 |
+(9,8) | → | 7 |
Automaton 2
final states:
{10}
transitions:
57 | → | 10 |
f7 | → | 12 |
f5 | → | 11 |
s(14) | → | 10 |
+(13,11) | → | 14 |
+(11,12) | → | 57 |
p(57) | → | 14 |
p(12) | → | 13 |