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)] | = | 
 
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| 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.