The rewrite relation of the following TRS is considered.
a(x1) | → | x1 | (1) |
a(a(x1)) | → | b(a(c(b(x1)))) | (2) |
b(x1) | → | c(x1) | (3) |
b(c(x1)) | → | a(x1) | (4) |
a(x1) | → | x1 | (1) |
a(a(x1)) | → | b(c(a(b(x1)))) | (5) |
b(x1) | → | c(x1) | (3) |
c(b(x1)) | → | a(x1) | (6) |
a#(a(x1)) | → | b#(x1) | (7) |
a#(a(x1)) | → | a#(b(x1)) | (8) |
a#(a(x1)) | → | c#(a(b(x1))) | (9) |
a#(a(x1)) | → | b#(c(a(b(x1)))) | (10) |
b#(x1) | → | c#(x1) | (11) |
c#(b(x1)) | → | a#(x1) | (12) |
[b#(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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[c(x1)] | = |
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a(x1) | → | x1 | (1) |
a(a(x1)) | → | b(c(a(b(x1)))) | (5) |
b(x1) | → | c(x1) | (3) |
c(b(x1)) | → | a(x1) | (6) |
a#(a(x1)) | → | b#(x1) | (7) |
c#(b(x1)) | → | a#(x1) | (12) |
The dependency pairs are split into 1 component.
a#(a(x1)) | → | a#(b(x1)) | (8) |
[b(x1)] | = |
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[a#(x1)] | = |
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[a(x1)] | = |
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[c(x1)] | = |
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a(x1) | → | x1 | (1) |
a(a(x1)) | → | b(c(a(b(x1)))) | (5) |
b(x1) | → | c(x1) | (3) |
c(b(x1)) | → | a(x1) | (6) |
a#(a(x1)) | → | a#(b(x1)) | (8) |
There are no pairs anymore.