The rewrite relation of the following TRS is considered.
a(a(x1)) | → | a(b(b(b(x1)))) | (1) |
b(a(x1)) | → | b(b(c(x1))) | (2) |
a(b(b(c(x1)))) | → | a(a(a(b(x1)))) | (3) |
b#(a(x1)) | → | b#(c(x1)) | (4) |
b#(a(x1)) | → | b#(b(c(x1))) | (5) |
a#(b(b(c(x1)))) | → | b#(x1) | (6) |
a#(b(b(c(x1)))) | → | a#(b(x1)) | (7) |
a#(b(b(c(x1)))) | → | a#(a(b(x1))) | (8) |
a#(b(b(c(x1)))) | → | a#(a(a(b(x1)))) | (9) |
a#(a(x1)) | → | b#(x1) | (10) |
a#(a(x1)) | → | b#(b(x1)) | (11) |
a#(a(x1)) | → | b#(b(b(x1))) | (12) |
a#(a(x1)) | → | a#(b(b(b(x1)))) | (13) |
[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(a(x1)) | → | a(b(b(b(x1)))) | (1) |
b(a(x1)) | → | b(b(c(x1))) | (2) |
a(b(b(c(x1)))) | → | a(a(a(b(x1)))) | (3) |
a#(b(b(c(x1)))) | → | b#(x1) | (6) |
a#(a(x1)) | → | b#(x1) | (10) |
a#(a(x1)) | → | b#(b(x1)) | (11) |
a#(a(x1)) | → | b#(b(b(x1))) | (12) |
The dependency pairs are split into 1 component.
a#(b(b(c(x1)))) | → | a#(b(x1)) | (7) |
a#(b(b(c(x1)))) | → | a#(a(b(x1))) | (8) |
a#(b(b(c(x1)))) | → | a#(a(a(b(x1)))) | (9) |
a#(a(x1)) | → | a#(b(b(b(x1)))) | (13) |
[c(x1)] | = |
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[b(x1)] | = |
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[a(x1)] | = |
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[a#(x1)] | = |
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a(a(x1)) | → | a(b(b(b(x1)))) | (1) |
b(a(x1)) | → | b(b(c(x1))) | (2) |
a(b(b(c(x1)))) | → | a(a(a(b(x1)))) | (3) |
a#(a(x1)) | → | a#(b(b(b(x1)))) | (13) |
The dependency pairs are split into 1 component.
a#(b(b(c(x1)))) | → | a#(b(x1)) | (7) |
a#(b(b(c(x1)))) | → | a#(a(b(x1))) | (8) |
a#(b(b(c(x1)))) | → | a#(a(a(b(x1)))) | (9) |
[c(x1)] | = |
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[b(x1)] | = |
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[a(x1)] | = |
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[a#(x1)] | = |
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a(a(x1)) | → | a(b(b(b(x1)))) | (1) |
b(a(x1)) | → | b(b(c(x1))) | (2) |
a(b(b(c(x1)))) | → | a(a(a(b(x1)))) | (3) |
a#(b(b(c(x1)))) | → | a#(a(b(x1))) | (8) |
a#(b(b(c(x1)))) | → | a#(a(a(b(x1)))) | (9) |
The dependency pairs are split into 1 component.
a#(b(b(c(x1)))) | → | a#(b(x1)) | (7) |
[c(x1)] | = |
x1 +
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[b(x1)] | = |
x1 +
|
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[a(x1)] | = |
x1 +
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[a#(x1)] | = |
x1 +
|
b(a(x1)) | → | b(b(c(x1))) | (2) |
a#(b(b(c(x1)))) | → | a#(b(x1)) | (7) |
b(a(x1)) | → | b(b(c(x1))) | (2) |
The dependency pairs are split into 0 components.