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