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