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