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