The rewrite relation of the following TRS is considered.
| 0(1(2(3(4(5(1(x1))))))) | → | 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(x1))))))))))))))))))) | (1) |
| 0(1(2(3(4(5(1(x1))))))) | → | 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(0(1(2(3(4(5(x1))))))))))))))))))))))))) | (2) |
| 1(5(4(3(2(1(0(x1))))))) | → | 5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))) | (3) |
| 1(5(4(3(2(1(0(x1))))))) | → | 5(4(3(2(1(0(5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))))))))) | (4) |
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(x1) | (5) |
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(5(4(3(2(1(x1)))))) | (6) |
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(1(5(4(3(2(1(x1))))))) | (7) |
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(0(5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))))) | (8) |
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))) | (9) |
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(0(1(1(5(4(3(2(1(x1))))))))) | (10) |
The dependency pairs are split into 1 component.
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(x1) | (5) |
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(5(4(3(2(1(x1)))))) | (6) |
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(1(5(4(3(2(1(x1))))))) | (7) |
| [5(x1)] | = |
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| [4(x1)] | = |
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| [3(x1)] | = |
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| [2(x1)] | = |
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| [1(x1)] | = |
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| [0(x1)] | = |
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| [1#(x1)] | = |
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| 1(5(4(3(2(1(0(x1))))))) | → | 5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))) | (3) |
| 1(5(4(3(2(1(0(x1))))))) | → | 5(4(3(2(1(0(5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))))))))) | (4) |
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(5(4(3(2(1(x1)))))) | (6) |
The dependency pairs are split into 1 component.
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(x1) | (5) |
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(1(5(4(3(2(1(x1))))))) | (7) |
| [5(x1)] | = |
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| [4(x1)] | = |
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| [3(x1)] | = |
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| [2(x1)] | = |
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| [1(x1)] | = |
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| [0(x1)] | = |
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| [1#(x1)] | = |
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| 1(5(4(3(2(1(0(x1))))))) | → | 5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))) | (3) |
| 1(5(4(3(2(1(0(x1))))))) | → | 5(4(3(2(1(0(5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))))))))) | (4) |
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(1(5(4(3(2(1(x1))))))) | (7) |
The dependency pairs are split into 1 component.
| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(x1) | (5) |
| [5(x1)] | = |
x1 +
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| [4(x1)] | = |
x1 +
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| [3(x1)] | = |
x1 +
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| [2(x1)] | = |
x1 +
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| [1(x1)] | = |
x1 +
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| [0(x1)] | = |
x1 +
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| [1#(x1)] | = |
x1 +
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| 1#(5(4(3(2(1(0(x1))))))) | → | 1#(x1) | (5) |
The dependency pairs are split into 0 components.