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