YES
by ttt2 (version ttt2 1.15)
The rewrite relation of the following TRS is considered.
| a(x0) | → | x0 |
| a(b(x0)) | → | b(a(c(a(x0)))) |
| b(x0) | → | x0 |
| c(c(c(x0))) | → | b(x0) |
| a#(b(x0)) | → | a#(x0) |
| a#(b(x0)) | → | c#(a(x0)) |
| a#(b(x0)) | → | a#(c(a(x0))) |
| a#(b(x0)) | → | b#(a(c(a(x0)))) |
| c#(c(c(x0))) | → | b#(x0) |
The dependency pairs are split into 1 component.
| a#(b(x0)) | → | a#(c(a(x0))) |
| a#(b(x0)) | → | a#(x0) |
| [b(x1)] | = |
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| [a#(x1)] | = |
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| [a(x1)] | = |
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| [c(x1)] | = |
|
| a(x0) | → | x0 |
| a(b(x0)) | → | b(a(c(a(x0)))) |
| b(x0) | → | x0 |
| c(c(c(x0))) | → | b(x0) |
| a#(b(x0)) | → | a#(c(a(x0))) |
| [b(x1)] | = |
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| [a#(x1)] | = |
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| [a(x1)] | = |
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| [c(x1)] | = |
|
| a(x0) | → | x0 |
| a(b(x0)) | → | b(a(c(a(x0)))) |
| b(x0) | → | x0 |
| c(c(c(x0))) | → | b(x0) |
There are no pairs anymore.