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#(c(x1)) | (3) |
| a#(a(x1)) | → | b#(b(c(x1))) | (4) |
| a#(a(x1)) | → | b#(b(b(c(x1)))) | (5) |
| b#(c(b(x1))) | → | b#(c(x1)) | (6) |
| b#(c(b(x1))) | → | a#(b(c(x1))) | (7) |
The dependency pairs are split into 1 component.
| b#(c(b(x1))) | → | b#(c(x1)) | (6) |
| b#(c(b(x1))) | → | a#(b(c(x1))) | (7) |
| a#(a(x1)) | → | b#(c(x1)) | (3) |
| [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)) | → | b(b(b(c(x1)))) | (1) |
| b(c(b(x1))) | → | a(b(c(x1))) | (2) |
| a#(a(x1)) | → | b#(c(x1)) | (3) |
The dependency pairs are split into 1 component.
| b#(c(b(x1))) | → | b#(c(x1)) | (6) |
| π(b#) | = | { 1 } |
| π(c) | = | { 1 } |
| b#(c(b(x1))) | → | b#(c(x1)) | (6) |
There are no pairs anymore.