The rewrite relation of the following TRS is considered.
a(a(x1)) | → | b(c(x1)) | (1) |
b(b(x1)) | → | c(d(x1)) | (2) |
c(c(x1)) | → | d(d(d(x1))) | (3) |
d(c(x1)) | → | b(f(x1)) | (4) |
d(d(d(x1))) | → | a(c(x1)) | (5) |
f(f(x1)) | → | f(b(x1)) | (6) |
b#(b(x1)) | → | d#(x1) | (7) |
d#(d(d(x1))) | → | c#(x1) | (8) |
d#(c(x1)) | → | f#(x1) | (9) |
c#(c(x1)) | → | d#(x1) | (10) |
a#(a(x1)) | → | c#(x1) | (11) |
d#(c(x1)) | → | b#(f(x1)) | (12) |
c#(c(x1)) | → | d#(d(d(x1))) | (13) |
a#(a(x1)) | → | b#(c(x1)) | (14) |
d#(d(d(x1))) | → | a#(c(x1)) | (15) |
f#(f(x1)) | → | f#(b(x1)) | (16) |
b#(b(x1)) | → | c#(d(x1)) | (17) |
c#(c(x1)) | → | d#(d(x1)) | (18) |
f#(f(x1)) | → | b#(x1) | (19) |
The dependency pairs are split into 1 component.
f#(f(x1)) | → | b#(x1) | (19) |
d#(c(x1)) | → | b#(f(x1)) | (12) |
c#(c(x1)) | → | d#(d(x1)) | (18) |
b#(b(x1)) | → | c#(d(x1)) | (17) |
f#(f(x1)) | → | f#(b(x1)) | (16) |
a#(a(x1)) | → | c#(x1) | (11) |
c#(c(x1)) | → | d#(x1) | (10) |
d#(c(x1)) | → | f#(x1) | (9) |
d#(d(d(x1))) | → | c#(x1) | (8) |
d#(d(d(x1))) | → | a#(c(x1)) | (15) |
a#(a(x1)) | → | b#(c(x1)) | (14) |
c#(c(x1)) | → | d#(d(d(x1))) | (13) |
b#(b(x1)) | → | d#(x1) | (7) |
[a(x1)] | = | x1 + 40 |
[d(x1)] | = | x1 + 28 |
[b(x1)] | = | x1 + 36 |
[c(x1)] | = | x1 + 44 |
[f(x1)] | = | x1 + 36 |
[d#(x1)] | = | x1 + 0 |
[f#(x1)] | = | x1 + 43 |
[c#(x1)] | = | x1 + 13 |
[a#(x1)] | = | x1 + 11 |
[b#(x1)] | = | x1 + 6 |
d(c(x1)) | → | b(f(x1)) | (4) |
a(a(x1)) | → | b(c(x1)) | (1) |
c(c(x1)) | → | d(d(d(x1))) | (3) |
d(d(d(x1))) | → | a(c(x1)) | (5) |
f(f(x1)) | → | f(b(x1)) | (6) |
b(b(x1)) | → | c(d(x1)) | (2) |
f#(f(x1)) | → | b#(x1) | (19) |
d#(c(x1)) | → | b#(f(x1)) | (12) |
c#(c(x1)) | → | d#(d(x1)) | (18) |
b#(b(x1)) | → | c#(d(x1)) | (17) |
a#(a(x1)) | → | c#(x1) | (11) |
c#(c(x1)) | → | d#(x1) | (10) |
d#(c(x1)) | → | f#(x1) | (9) |
d#(d(d(x1))) | → | c#(x1) | (8) |
d#(d(d(x1))) | → | a#(c(x1)) | (15) |
a#(a(x1)) | → | b#(c(x1)) | (14) |
c#(c(x1)) | → | d#(d(d(x1))) | (13) |
b#(b(x1)) | → | d#(x1) | (7) |
The dependency pairs are split into 1 component.
f#(f(x1)) | → | f#(b(x1)) | (16) |
[a(x1)] | = | 1 |
[d(x1)] | = | 1 |
[b(x1)] | = | 1 |
[c(x1)] | = | 1 |
[f(x1)] | = | 2 |
[d#(x1)] | = | x1 + 0 |
[f#(x1)] | = | x1 + 43 |
[c#(x1)] | = | x1 + 13 |
[a#(x1)] | = | x1 + 11 |
[b#(x1)] | = | x1 + 6 |
d(c(x1)) | → | b(f(x1)) | (4) |
a(a(x1)) | → | b(c(x1)) | (1) |
c(c(x1)) | → | d(d(d(x1))) | (3) |
d(d(d(x1))) | → | a(c(x1)) | (5) |
f(f(x1)) | → | f(b(x1)) | (6) |
b(b(x1)) | → | c(d(x1)) | (2) |
f#(f(x1)) | → | f#(b(x1)) | (16) |
The dependency pairs are split into 0 components.