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.