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.