The rewrite relation of the following TRS is considered.
a(a(x1)) | → | x1 | (1) |
a(b(x1)) | → | c(x1) | (2) |
c(c(x1)) | → | b(c(b(a(a(x1))))) | (3) |
a(a(x1)) | → | x1 | (1) |
b(a(x1)) | → | c(x1) | (4) |
c(c(x1)) | → | a(a(b(c(b(x1))))) | (5) |
c#(c(x1)) | → | c#(b(x1)) | (6) |
c#(c(x1)) | → | b#(x1) | (7) |
c#(c(x1)) | → | b#(c(b(x1))) | (8) |
c#(c(x1)) | → | a#(b(c(b(x1)))) | (9) |
c#(c(x1)) | → | a#(a(b(c(b(x1))))) | (10) |
b#(a(x1)) | → | c#(x1) | (11) |
[c(x1)] | = |
x1 +
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[b(x1)] | = |
x1 +
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[a(x1)] | = |
x1 +
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[c#(x1)] | = |
x1 +
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[b#(x1)] | = |
x1 +
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[a#(x1)] | = |
x1 +
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a(a(x1)) | → | x1 | (1) |
b(a(x1)) | → | c(x1) | (4) |
c(c(x1)) | → | a(a(b(c(b(x1))))) | (5) |
c#(c(x1)) | → | a#(b(c(b(x1)))) | (9) |
c#(c(x1)) | → | a#(a(b(c(b(x1))))) | (10) |
The dependency pairs are split into 1 component.
c#(c(x1)) | → | c#(b(x1)) | (6) |
c#(c(x1)) | → | b#(x1) | (7) |
b#(a(x1)) | → | c#(x1) | (11) |
c#(c(x1)) | → | b#(c(b(x1))) | (8) |
[c(x1)] | = |
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[b(x1)] | = |
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[a(x1)] | = |
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[c#(x1)] | = |
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[b#(x1)] | = |
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a(a(x1)) | → | x1 | (1) |
b(a(x1)) | → | c(x1) | (4) |
c(c(x1)) | → | a(a(b(c(b(x1))))) | (5) |
c#(c(x1)) | → | b#(c(b(x1))) | (8) |
The dependency pairs are split into 1 component.
c#(c(x1)) | → | c#(b(x1)) | (6) |
c#(c(x1)) | → | b#(x1) | (7) |
b#(a(x1)) | → | c#(x1) | (11) |
[c(x1)] | = |
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[b(x1)] | = |
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[a(x1)] | = |
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[c#(x1)] | = |
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[b#(x1)] | = |
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a(a(x1)) | → | x1 | (1) |
b(a(x1)) | → | c(x1) | (4) |
c(c(x1)) | → | a(a(b(c(b(x1))))) | (5) |
c#(c(x1)) | → | c#(b(x1)) | (6) |
The dependency pairs are split into 1 component.
c#(c(x1)) | → | b#(x1) | (7) |
b#(a(x1)) | → | c#(x1) | (11) |
[c(x1)] | = |
x1 +
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[a(x1)] | = |
x1 +
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[c#(x1)] | = |
x1 +
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[b#(x1)] | = |
x1 +
|
c#(c(x1)) | → | b#(x1) | (7) |
b#(a(x1)) | → | c#(x1) | (11) |
The dependency pairs are split into 0 components.