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