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