The rewrite relation of the following TRS is considered.
a(b(x1)) | → | b(c(a(x1))) | (1) |
b(c(x1)) | → | c(b(b(x1))) | (2) |
c(a(x1)) | → | a(c(x1)) | (3) |
b#(c(x1)) | → | b#(b(x1)) | (4) |
a#(b(x1)) | → | b#(c(a(x1))) | (5) |
a#(b(x1)) | → | c#(a(x1)) | (6) |
c#(a(x1)) | → | c#(x1) | (7) |
b#(c(x1)) | → | b#(x1) | (8) |
b#(c(x1)) | → | c#(b(b(x1))) | (9) |
a#(b(x1)) | → | a#(x1) | (10) |
c#(a(x1)) | → | a#(c(x1)) | (11) |
The dependency pairs are split into 1 component.
c#(a(x1)) | → | a#(c(x1)) | (11) |
a#(b(x1)) | → | a#(x1) | (10) |
b#(c(x1)) | → | c#(b(b(x1))) | (9) |
b#(c(x1)) | → | b#(x1) | (8) |
a#(b(x1)) | → | b#(c(a(x1))) | (5) |
c#(a(x1)) | → | c#(x1) | (7) |
a#(b(x1)) | → | c#(a(x1)) | (6) |
b#(c(x1)) | → | b#(b(x1)) | (4) |
[a(x1)] | = | x1 + 0 |
[b(x1)] | = | x1 + 3 |
[c(x1)] | = | 0 |
[c#(x1)] | = | 1 |
[a#(x1)] | = | x1 + 0 |
[b#(x1)] | = | 2 |
a(b(x1)) | → | b(c(a(x1))) | (1) |
c(a(x1)) | → | a(c(x1)) | (3) |
b(c(x1)) | → | c(b(b(x1))) | (2) |
c#(a(x1)) | → | a#(c(x1)) | (11) |
a#(b(x1)) | → | a#(x1) | (10) |
b#(c(x1)) | → | c#(b(b(x1))) | (9) |
a#(b(x1)) | → | b#(c(a(x1))) | (5) |
a#(b(x1)) | → | c#(a(x1)) | (6) |
The dependency pairs are split into 2 components.
c#(a(x1)) | → | c#(x1) | (7) |
[a(x1)] | = | x1 + 1 |
[b(x1)] | = | 40651 |
[c(x1)] | = | x1 + 0 |
[c#(x1)] | = | x1 + 1 |
[a#(x1)] | = | x1 + 0 |
[b#(x1)] | = | 2 |
a(b(x1)) | → | b(c(a(x1))) | (1) |
c(a(x1)) | → | a(c(x1)) | (3) |
b(c(x1)) | → | c(b(b(x1))) | (2) |
c#(a(x1)) | → | c#(x1) | (7) |
The dependency pairs are split into 0 components.
b#(c(x1)) | → | b#(x1) | (8) |
b#(c(x1)) | → | b#(b(x1)) | (4) |
[a(x1)] | = |
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[b(x1)] | = |
x1 +
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[c(x1)] | = |
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[c#(x1)] | = |
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[a#(x1)] | = |
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[b#(x1)] | = |
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a(b(x1)) | → | b(c(a(x1))) | (1) |
c(a(x1)) | → | a(c(x1)) | (3) |
b(c(x1)) | → | c(b(b(x1))) | (2) |
b#(c(x1)) | → | b#(x1) | (8) |
b#(c(x1)) | → | b#(b(x1)) | (4) |
The dependency pairs are split into 0 components.