We consider the TRS containing the following rules:
br(0,y,z) | → | y | (1) |
br(s(x),y,z) | → | z | (2) |
p(0) | → | 0 | (3) |
p(s(x)) | → | x | (4) |
+(x,y) | → | br(x,y,+(p(x),s(y))) | (5) |
+(x,y) | → | br(y,x,+(s(x),p(y))) | (6) |
The underlying signature is as follows:
{br/3, 0/0, s/1, p/1, +/2}t0 | = | +(0,c_1) |
→ | br(c_1,0,+(s(0),p(c_1))) | |
→ | br(c_1,0,br(s(0),p(c_1),+(p(s(0)),s(p(c_1))))) | |
→ | br(c_1,0,+(p(s(0)),s(p(c_1)))) | |
→ | br(c_1,0,+(0,s(p(c_1)))) | |
= | t4 |
t0 | = | +(0,c_1) |
→ | br(0,c_1,+(p(0),s(c_1))) | |
→ | c_1 | |
= | t2 |