The rewrite relation of the following TRS is considered.
a(p(x1)) | → | p(a(A(x1))) | (1) |
a(A(x1)) | → | A(a(x1)) | (2) |
p(A(A(x1))) | → | a(p(x1)) | (3) |
p#(A(A(x1))) | → | a#(p(x1)) | (4) |
a#(p(x1)) | → | a#(A(x1)) | (5) |
p#(A(A(x1))) | → | p#(x1) | (6) |
a#(A(x1)) | → | a#(x1) | (7) |
a#(p(x1)) | → | p#(a(A(x1))) | (8) |
The dependency pairs are split into 1 component.
a#(p(x1)) | → | p#(a(A(x1))) | (8) |
a#(A(x1)) | → | a#(x1) | (7) |
p#(A(A(x1))) | → | p#(x1) | (6) |
a#(p(x1)) | → | a#(A(x1)) | (5) |
p#(A(A(x1))) | → | a#(p(x1)) | (4) |
[a(x1)] | = | x1 + 0 |
[p#(x1)] | = | x1 + 1 |
[p(x1)] | = | x1 + 1 |
[A(x1)] | = | x1 + 0 |
[a#(x1)] | = | x1 + 0 |
a(p(x1)) | → | p(a(A(x1))) | (1) |
p(A(A(x1))) | → | a(p(x1)) | (3) |
a(A(x1)) | → | A(a(x1)) | (2) |
a#(p(x1)) | → | a#(A(x1)) | (5) |
The dependency pairs are split into 1 component.
a#(p(x1)) | → | p#(a(A(x1))) | (8) |
p#(A(A(x1))) | → | p#(x1) | (6) |
p#(A(A(x1))) | → | a#(p(x1)) | (4) |
a#(A(x1)) | → | a#(x1) | (7) |
[a(x1)] | = | x1 + 0 |
[p#(x1)] | = | 1 |
[p(x1)] | = | 1 |
[A(x1)] | = | x1 + 32286 |
[a#(x1)] | = | x1 + 0 |
a(p(x1)) | → | p(a(A(x1))) | (1) |
p(A(A(x1))) | → | a(p(x1)) | (3) |
a(A(x1)) | → | A(a(x1)) | (2) |
a#(A(x1)) | → | a#(x1) | (7) |
The dependency pairs are split into 1 component.
a#(p(x1)) | → | p#(a(A(x1))) | (8) |
p#(A(A(x1))) | → | p#(x1) | (6) |
p#(A(A(x1))) | → | a#(p(x1)) | (4) |
[a(x1)] | = |
x1 +
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[p#(x1)] | = |
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[p(x1)] | = |
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[A(x1)] | = |
x1 +
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[a#(x1)] | = |
x1 +
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a(p(x1)) | → | p(a(A(x1))) | (1) |
p(A(A(x1))) | → | a(p(x1)) | (3) |
a(A(x1)) | → | A(a(x1)) | (2) |
a#(p(x1)) | → | p#(a(A(x1))) | (8) |
p#(A(A(x1))) | → | p#(x1) | (6) |
p#(A(A(x1))) | → | a#(p(x1)) | (4) |
The dependency pairs are split into 0 components.