The rewrite relation of the following TRS is considered.
b(d(b(x1))) | → | c(d(b(x1))) | (1) |
b(a(c(x1))) | → | b(c(x1)) | (2) |
a(d(x1)) | → | d(c(x1)) | (3) |
b(b(b(x1))) | → | a(b(c(x1))) | (4) |
d(c(x1)) | → | b(d(x1)) | (5) |
d(c(x1)) | → | d(b(d(x1))) | (6) |
d(a(c(x1))) | → | b(b(x1)) | (7) |
[b(x1)] | = | 1 · x1 + 1 |
[d(x1)] | = | 1 · x1 |
[c(x1)] | = | 1 · x1 + 1 |
[a(x1)] | = | 1 · x1 + 1 |
b(a(c(x1))) | → | b(c(x1)) | (2) |
a#(d(x1)) | → | d#(c(x1)) | (8) |
b#(b(b(x1))) | → | a#(b(c(x1))) | (9) |
b#(b(b(x1))) | → | b#(c(x1)) | (10) |
d#(c(x1)) | → | b#(d(x1)) | (11) |
d#(c(x1)) | → | d#(x1) | (12) |
d#(c(x1)) | → | d#(b(d(x1))) | (13) |
d#(a(c(x1))) | → | b#(b(x1)) | (14) |
d#(a(c(x1))) | → | b#(x1) | (15) |
The dependency pairs are split into 1 component.
d#(c(x1)) | → | d#(b(d(x1))) | (13) |
d#(c(x1)) | → | d#(x1) | (12) |
[d(x1)] | = | 1 · x1 |
[c(x1)] | = | 1 · x1 |
[b(x1)] | = | 1 · x1 |
[a(x1)] | = | 1 · x1 |
[d#(x1)] | = | 1 · x1 |
d(c(x1)) | → | b(d(x1)) | (5) |
d(c(x1)) | → | d(b(d(x1))) | (6) |
d(a(c(x1))) | → | b(b(x1)) | (7) |
b(d(b(x1))) | → | c(d(b(x1))) | (1) |
b(b(b(x1))) | → | a(b(c(x1))) | (4) |
[d#(x1)] | = | 1 · x1 |
[c(x1)] | = | 1 + 1 · x1 |
[b(x1)] | = | 1 + 1 · x1 |
[d(x1)] | = | 1 · x1 |
[a(x1)] | = | 1 + 1 · x1 |
d#(c(x1)) | → | d#(x1) | (12) |
[d#(x1)] | = |
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[c(x1)] | = |
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[b(x1)] | = |
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[d(x1)] | = |
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[a(x1)] | = |
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d#(c(x1)) | → | d#(b(d(x1))) | (13) |
There are no pairs anymore.