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