The rewrite relation of the following TRS is considered.
a(b(x1)) | → | b(b(a(x1))) | (1) |
b(c(x1)) | → | c(b(b(x1))) | (2) |
c(a(x1)) | → | a(c(c(x1))) | (3) |
b#(c(x1)) | → | b#(b(x1)) | (4) |
c#(a(x1)) | → | c#(x1) | (5) |
a#(b(x1)) | → | a#(x1) | (6) |
a#(b(x1)) | → | b#(b(a(x1))) | (7) |
c#(a(x1)) | → | c#(c(x1)) | (8) |
b#(c(x1)) | → | b#(x1) | (9) |
b#(c(x1)) | → | c#(b(b(x1))) | (10) |
a#(b(x1)) | → | b#(a(x1)) | (11) |
c#(a(x1)) | → | a#(c(c(x1))) | (12) |
The dependency pairs are split into 1 component.
c#(a(x1)) | → | a#(c(c(x1))) | (12) |
a#(b(x1)) | → | b#(a(x1)) | (11) |
b#(c(x1)) | → | c#(b(b(x1))) | (10) |
b#(c(x1)) | → | b#(x1) | (9) |
a#(b(x1)) | → | a#(x1) | (6) |
c#(a(x1)) | → | c#(x1) | (5) |
c#(a(x1)) | → | c#(c(x1)) | (8) |
a#(b(x1)) | → | b#(b(a(x1))) | (7) |
b#(c(x1)) | → | b#(b(x1)) | (4) |
[a(x1)] | = | 1 |
[b(x1)] | = | x1 + 0 |
[c(x1)] | = | x1 + 4 |
[c#(x1)] | = | 3 |
[a#(x1)] | = | 2 |
[b#(x1)] | = | x1 + 0 |
a(b(x1)) | → | b(b(a(x1))) | (1) |
c(a(x1)) | → | a(c(c(x1))) | (3) |
b(c(x1)) | → | c(b(b(x1))) | (2) |
c#(a(x1)) | → | a#(c(c(x1))) | (12) |
a#(b(x1)) | → | b#(a(x1)) | (11) |
b#(c(x1)) | → | c#(b(b(x1))) | (10) |
b#(c(x1)) | → | b#(x1) | (9) |
a#(b(x1)) | → | b#(b(a(x1))) | (7) |
b#(c(x1)) | → | b#(b(x1)) | (4) |
The dependency pairs are split into 2 components.
a#(b(x1)) | → | a#(x1) | (6) |
[a(x1)] | = | 0 |
[b(x1)] | = | x1 + 1 |
[c(x1)] | = | 0 |
[c#(x1)] | = | 0 |
[a#(x1)] | = | x1 + 0 |
[b#(x1)] | = | 0 |
a#(b(x1)) | → | a#(x1) | (6) |
The dependency pairs are split into 0 components.
c#(a(x1)) | → | c#(x1) | (5) |
c#(a(x1)) | → | c#(c(x1)) | (8) |
[a(x1)] | = | x1 + 1 |
[b(x1)] | = | 41063 |
[c(x1)] | = | x1 + 0 |
[c#(x1)] | = | x1 + 3 |
[a#(x1)] | = | 2 |
[b#(x1)] | = | x1 + 0 |
a(b(x1)) | → | b(b(a(x1))) | (1) |
c(a(x1)) | → | a(c(c(x1))) | (3) |
b(c(x1)) | → | c(b(b(x1))) | (2) |
c#(a(x1)) | → | c#(x1) | (5) |
c#(a(x1)) | → | c#(c(x1)) | (8) |
The dependency pairs are split into 0 components.