The rewrite relation of the following TRS is considered.
a(b(x1)) | → | x1 | (1) |
a(c(x1)) | → | b(c(a(a(x1)))) | (2) |
c(b(x1)) | → | a(c(x1)) | (3) |
a#(c(x1)) | → | a#(x1) | (4) |
a#(c(x1)) | → | a#(a(x1)) | (5) |
a#(c(x1)) | → | c#(a(a(x1))) | (6) |
c#(b(x1)) | → | c#(x1) | (7) |
c#(b(x1)) | → | a#(c(x1)) | (8) |
π(c#) | = | { 1, 1 } |
π(a#) | = | { 1 } |
π(c) | = | { 1, 1 } |
π(a) | = | { 1 } |
π(b) | = | { 1 } |
a#(c(x1)) | → | a#(x1) | (4) |
a#(c(x1)) | → | a#(a(x1)) | (5) |
[c#(x1)] | = |
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[a(x1)] | = |
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[a#(x1)] | = |
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[b(x1)] | = |
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[c(x1)] | = |
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a(b(x1)) | → | x1 | (1) |
a(c(x1)) | → | b(c(a(a(x1)))) | (2) |
c(b(x1)) | → | a(c(x1)) | (3) |
c#(b(x1)) | → | c#(x1) | (7) |
c#(b(x1)) | → | a#(c(x1)) | (8) |
The dependency pairs are split into 0 components.