The rewrite relation of the following TRS is considered.
a(x1) | → | x1 | (1) |
a(b(b(x1))) | → | b(b(b(c(x1)))) | (2) |
c(b(x1)) | → | a(a(x1)) | (3) |
a(x1) | → | x1 | (1) |
b(b(a(x1))) | → | c(b(b(b(x1)))) | (4) |
b(c(x1)) | → | a(a(x1)) | (5) |
b#(b(a(x1))) | → | b#(x1) | (6) |
b#(b(a(x1))) | → | b#(b(x1)) | (7) |
b#(b(a(x1))) | → | b#(b(b(x1))) | (8) |
b#(c(x1)) | → | a#(x1) | (9) |
b#(c(x1)) | → | a#(a(x1)) | (10) |
The dependency pairs are split into 1 component.
b#(b(a(x1))) | → | b#(b(b(x1))) | (8) |
b#(b(a(x1))) | → | b#(x1) | (6) |
b#(b(a(x1))) | → | b#(b(x1)) | (7) |
[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) | → | x1 | (1) |
b(b(a(x1))) | → | c(b(b(b(x1)))) | (4) |
b(c(x1)) | → | a(a(x1)) | (5) |
b#(b(a(x1))) | → | b#(x1) | (6) |
[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) | → | x1 | (1) |
b(b(a(x1))) | → | c(b(b(b(x1)))) | (4) |
b(c(x1)) | → | a(a(x1)) | (5) |
b#(b(a(x1))) | → | b#(b(b(x1))) | (8) |
[b#(x1)] | = | 1/2 · x1 + 2 |
[b(x1)] | = | 2 · x1 + 0 |
[a(x1)] | = | 1 · x1 + 1/2 |
[c(x1)] | = | 1/2 · x1 + 2 |
a(x1) | → | x1 | (1) |
b(b(a(x1))) | → | c(b(b(b(x1)))) | (4) |
b(c(x1)) | → | a(a(x1)) | (5) |
b#(b(a(x1))) | → | b#(b(x1)) | (7) |
There are no pairs anymore.