The rewrite relation of the following TRS is considered.
b(b(a(b(b(a(b(b(b(b(x1)))))))))) | → | b(b(b(b(b(a(b(b(a(b(b(a(b(x1))))))))))))) | (1) |
b(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b(a(b(b(a(b(b(a(b(b(b(b(b(x1))))))))))))) | (2) |
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(a(b(b(a(b(b(a(b(b(b(b(b(x1))))))))))))) | (3) |
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(b(a(b(b(a(b(b(b(b(b(x1))))))))))) | (4) |
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(a(b(b(a(b(b(b(b(b(x1)))))))))) | (5) |
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(b(a(b(b(b(b(b(x1)))))))) | (6) |
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(a(b(b(b(b(b(x1))))))) | (7) |
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(b(b(b(b(x1))))) | (8) |
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(b(b(b(x1)))) | (9) |
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(b(b(x1))) | (10) |
The dependency pairs are split into 1 component.
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(b(b(b(x1)))) | (9) |
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(b(b(b(b(x1))))) | (8) |
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(b(b(x1))) | (10) |
[b#(x1)] | = |
|
||||||||||||||||||
[b(x1)] | = |
|
||||||||||||||||||
[a(x1)] | = |
|
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(b(b(x1))) | (10) |
[b#(x1)] | = |
|
||||||||||||||||||
[b(x1)] | = |
|
||||||||||||||||||
[a(x1)] | = |
|
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(b(b(b(b(x1))))) | (8) |
[b#(x1)] | = | 0 + 1 · x1 |
[b(x1)] | = | 4 + 2 · x1 |
[a(x1)] | = | 0 + 1/4 · x1 |
b#(b(b(b(a(b(b(a(b(b(x1)))))))))) | → | b#(b(b(b(x1)))) | (9) |
There are no pairs anymore.