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.