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