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