The rewrite relation of the following TRS is considered.
| c(c(x1)) | → | a(b(x1)) | (1) |
| b(x1) | → | a(a(x1)) | (2) |
| b(b(b(x1))) | → | a(c(b(x1))) | (3) |
| a(c(a(x1))) | → | a(c(c(x1))) | (4) |
| c#(c(x1)) | → | a#(b(x1)) | (5) |
| c#(c(x1)) | → | b#(x1) | (6) |
| b#(x1) | → | a#(a(x1)) | (7) |
| b#(x1) | → | a#(x1) | (8) |
| b#(b(b(x1))) | → | a#(c(b(x1))) | (9) |
| b#(b(b(x1))) | → | c#(b(x1)) | (10) |
| a#(c(a(x1))) | → | a#(c(c(x1))) | (11) |
| a#(c(a(x1))) | → | c#(c(x1)) | (12) |
| a#(c(a(x1))) | → | c#(x1) | (13) |
The dependency pairs are split into 1 component.
| c#(c(x1)) | → | b#(x1) | (6) |
| b#(x1) | → | a#(x1) | (8) |
| a#(c(a(x1))) | → | a#(c(c(x1))) | (11) |
| a#(c(a(x1))) | → | c#(c(x1)) | (12) |
| a#(c(a(x1))) | → | c#(x1) | (13) |
| b#(b(b(x1))) | → | a#(c(b(x1))) | (9) |
| [c#(x1)] | = |
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| [c(x1)] | = |
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| [b#(x1)] | = |
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| [a#(x1)] | = |
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| [a(x1)] | = |
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| [b(x1)] | = |
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| b#(x1) | → | a#(x1) | (8) |
| [c#(x1)] | = |
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| [c(x1)] | = |
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| [b#(x1)] | = |
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| [a#(x1)] | = |
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| [a(x1)] | = |
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| [b(x1)] | = |
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| b#(b(b(x1))) | → | a#(c(b(x1))) | (9) |
The dependency pairs are split into 1 component.
| a#(c(a(x1))) | → | a#(c(c(x1))) | (11) |
| [a#(x1)] | = |
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| [c(x1)] | = |
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| [a(x1)] | = |
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| [b(x1)] | = |
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| a#(c(a(x1))) | → | a#(c(c(x1))) | (11) |
There are no pairs anymore.