YES
by ttt2 (version ttt2 1.15)
The rewrite relation of the following TRS is considered.
| a(a(x0)) | → | b(a(c(b(x0)))) | 
| b(b(x0)) | → | a(a(x0)) | 
| c(a(x0)) | → | x0 | 
| a#(a(x0)) | → | b#(x0) | 
| a#(a(x0)) | → | c#(b(x0)) | 
| a#(a(x0)) | → | a#(c(b(x0))) | 
| a#(a(x0)) | → | b#(a(c(b(x0)))) | 
| b#(b(x0)) | → | a#(x0) | 
| b#(b(x0)) | → | a#(a(x0)) | 
The dependency pairs are split into 1 component.
| b#(b(x0)) | → | a#(a(x0)) | 
| a#(a(x0)) | → | b#(x0) | 
| b#(b(x0)) | → | a#(x0) | 
| a#(a(x0)) | → | a#(c(b(x0))) | 
| a#(a(x0)) | → | b#(a(c(b(x0)))) | 
| [b#(x1)] | = | 9 · x1 + 14 | 
| [b(x1)] | = | 4 · x1 + 5 | 
| [a#(x1)] | = | 9 · x1 + 0 | 
| [a(x1)] | = | 4 · x1 + 5 | 
| [c(x1)] | = | -4 · x1 + 0 | 
| a(a(x0)) | → | b(a(c(b(x0)))) | 
| b(b(x0)) | → | a(a(x0)) | 
| c(a(x0)) | → | x0 | 
| b#(b(x0)) | → | a#(a(x0)) | 
| a#(a(x0)) | → | b#(x0) | 
| a#(a(x0)) | → | b#(a(c(b(x0)))) | 
| [b#(x1)] | = | 
 
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| [b(x1)] | = | 
 
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| [a#(x1)] | = | 
 
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| [a(x1)] | = | 
 
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| [c(x1)] | = | 
 
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| a(a(x0)) | → | b(a(c(b(x0)))) | 
| b(b(x0)) | → | a(a(x0)) | 
| c(a(x0)) | → | x0 | 
| b#(b(x0)) | → | a#(a(x0)) | 
| a#(a(x0)) | → | b#(a(c(b(x0)))) | 
| [b#(x1)] | = | 
 
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| [b(x1)] | = | 
 
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| [a#(x1)] | = | 
 
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| [a(x1)] | = | 
 
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| [c(x1)] | = | 
 
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| a(a(x0)) | → | b(a(c(b(x0)))) | 
| b(b(x0)) | → | a(a(x0)) | 
| c(a(x0)) | → | x0 | 
| a#(a(x0)) | → | b#(a(c(b(x0)))) | 
The dependency pairs are split into 0 components.