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