The rewrite relation of the following TRS is considered.
| a(a(x1)) | → | b(b(x1)) | (1) |
| c(c(b(x1))) | → | d(c(a(x1))) | (2) |
| a(x1) | → | d(c(c(x1))) | (3) |
| c(d(x1)) | → | b(c(x1)) | (4) |
| a(a(x1)) | → | b(b(x1)) | (1) |
| b(c(c(x1))) | → | a(c(d(x1))) | (5) |
| a(x1) | → | c(c(d(x1))) | (6) |
| d(c(x1)) | → | c(b(x1)) | (7) |
| a#(a(x1)) | → | b#(b(x1)) | (8) |
| a#(a(x1)) | → | b#(x1) | (9) |
| b#(c(c(x1))) | → | a#(c(d(x1))) | (10) |
| b#(c(c(x1))) | → | d#(x1) | (11) |
| a#(x1) | → | d#(x1) | (12) |
| d#(c(x1)) | → | b#(x1) | (13) |
The dependency pairs are split into 1 component.
| a#(x1) | → | d#(x1) | (12) |
| d#(c(x1)) | → | b#(x1) | (13) |
| b#(c(c(x1))) | → | a#(c(d(x1))) | (10) |
| b#(c(c(x1))) | → | d#(x1) | (11) |
| [d(x1)] | = | 1 · x1 |
| [c(x1)] | = | 1 · x1 |
| [b(x1)] | = | 1 · x1 |
| [a(x1)] | = | 1 · x1 |
| [d#(x1)] | = | 1 · x1 |
| [a#(x1)] | = | 1 · x1 |
| [b#(x1)] | = | 1 · x1 |
| d(c(x1)) | → | c(b(x1)) | (7) |
| b(c(c(x1))) | → | a(c(d(x1))) | (5) |
| a(x1) | → | c(c(d(x1))) | (6) |
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Hence, it suffices to show innermost termination in the following.| [a#(x1)] | = |
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| [d#(x1)] | = |
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| [c(x1)] | = |
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| [b#(x1)] | = |
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| [d(x1)] | = |
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| [b(x1)] | = |
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| [a(x1)] | = |
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| b#(c(c(x1))) | → | d#(x1) | (11) |
| [a#(x1)] | = |
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| [d#(x1)] | = |
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| [c(x1)] | = |
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| [b#(x1)] | = |
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| [d(x1)] | = |
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| [b(x1)] | = |
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| [a(x1)] | = |
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| a#(x1) | → | d#(x1) | (12) |
The dependency pairs are split into 0 components.