The rewrite relation of the following TRS is considered.
a(x1) | → | b(c(x1)) | (1) |
a(b(x1)) | → | b(a(x1)) | (2) |
d(c(x1)) | → | d(a(x1)) | (3) |
a(c(x1)) | → | c(a(x1)) | (4) |
d#(c(x1)) | → | d#(a(x1)) | (5) |
d#(c(x1)) | → | a#(x1) | (6) |
a#(b(x1)) | → | a#(x1) | (7) |
a#(c(x1)) | → | a#(x1) | (8) |
The dependency pairs are split into 2 components.
d#(c(x1)) | → | d#(a(x1)) | (5) |
[a(x1)] | = | x1 + 1 |
[d(x1)] | = | 0 |
[b(x1)] | = | 1 |
[c(x1)] | = | x1 + 2 |
[d#(x1)] | = | x1 + 0 |
[a#(x1)] | = | 0 |
a(c(x1)) | → | c(a(x1)) | (4) |
a(x1) | → | b(c(x1)) | (1) |
a(b(x1)) | → | b(a(x1)) | (2) |
d#(c(x1)) | → | d#(a(x1)) | (5) |
The dependency pairs are split into 0 components.
a#(c(x1)) | → | a#(x1) | (8) |
a#(b(x1)) | → | a#(x1) | (7) |
[a(x1)] | = | x1 + 1 |
[d(x1)] | = | 0 |
[b(x1)] | = | x1 + 1 |
[c(x1)] | = | x1 + 0 |
[d#(x1)] | = | x1 + 0 |
[a#(x1)] | = | x1 + 0 |
a(c(x1)) | → | c(a(x1)) | (4) |
a(x1) | → | b(c(x1)) | (1) |
a(b(x1)) | → | b(a(x1)) | (2) |
a#(b(x1)) | → | a#(x1) | (7) |
The dependency pairs are split into 1 component.
a#(c(x1)) | → | a#(x1) | (8) |
[a(x1)] | = | x1 + 1 |
[d(x1)] | = | 0 |
[b(x1)] | = | 1 |
[c(x1)] | = | x1 + 7720 |
[d#(x1)] | = | x1 + 0 |
[a#(x1)] | = | x1 + 0 |
a(c(x1)) | → | c(a(x1)) | (4) |
a(x1) | → | b(c(x1)) | (1) |
a(b(x1)) | → | b(a(x1)) | (2) |
a#(c(x1)) | → | a#(x1) | (8) |
The dependency pairs are split into 0 components.