The rewrite relation of the following TRS is considered.
c(c(a(x1))) | → | a(a(c(x1))) | (1) |
b(b(a(x1))) | → | b(b(b(x1))) | (2) |
b(c(b(x1))) | → | c(b(c(x1))) | (3) |
b(a(a(x1))) | → | c(a(c(x1))) | (4) |
c(a(a(x1))) | → | a(a(c(x1))) | (5) |
c(c(b(x1))) | → | c(b(a(x1))) | (6) |
a(c(c(x1))) | → | c(a(a(x1))) | (7) |
a(b(b(x1))) | → | b(b(b(x1))) | (8) |
b(c(b(x1))) | → | c(b(c(x1))) | (3) |
a(a(b(x1))) | → | c(a(c(x1))) | (9) |
a(a(c(x1))) | → | c(a(a(x1))) | (10) |
b(c(c(x1))) | → | a(b(c(x1))) | (11) |
b#(c(c(x1))) | → | b#(c(x1)) | (12) |
b#(c(c(x1))) | → | a#(b(c(x1))) | (13) |
b#(c(b(x1))) | → | b#(c(x1)) | (14) |
a#(c(c(x1))) | → | a#(x1) | (15) |
a#(c(c(x1))) | → | a#(a(x1)) | (16) |
a#(b(b(x1))) | → | b#(b(b(x1))) | (17) |
a#(a(c(x1))) | → | a#(x1) | (18) |
a#(a(c(x1))) | → | a#(a(x1)) | (19) |
a#(a(b(x1))) | → | a#(c(x1)) | (20) |
[c(x1)] | = |
x1 +
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[b(x1)] | = |
x1 +
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[a(x1)] | = |
x1 +
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[b#(x1)] | = |
x1 +
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[a#(x1)] | = |
x1 +
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a(c(c(x1))) | → | c(a(a(x1))) | (7) |
a(b(b(x1))) | → | b(b(b(x1))) | (8) |
b(c(b(x1))) | → | c(b(c(x1))) | (3) |
a(a(b(x1))) | → | c(a(c(x1))) | (9) |
a(a(c(x1))) | → | c(a(a(x1))) | (10) |
b(c(c(x1))) | → | a(b(c(x1))) | (11) |
b#(c(c(x1))) | → | b#(c(x1)) | (12) |
b#(c(b(x1))) | → | b#(c(x1)) | (14) |
a#(c(c(x1))) | → | a#(x1) | (15) |
a#(c(c(x1))) | → | a#(a(x1)) | (16) |
a#(a(c(x1))) | → | a#(x1) | (18) |
a#(a(c(x1))) | → | a#(a(x1)) | (19) |
a#(a(b(x1))) | → | a#(c(x1)) | (20) |
The dependency pairs are split into 1 component.
b#(c(c(x1))) | → | a#(b(c(x1))) | (13) |
a#(b(b(x1))) | → | b#(b(b(x1))) | (17) |
[c(x1)] | = |
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[b(x1)] | = |
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[a(x1)] | = |
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[b#(x1)] | = |
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[a#(x1)] | = |
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a(c(c(x1))) | → | c(a(a(x1))) | (7) |
a(b(b(x1))) | → | b(b(b(x1))) | (8) |
b(c(b(x1))) | → | c(b(c(x1))) | (3) |
a(a(b(x1))) | → | c(a(c(x1))) | (9) |
a(a(c(x1))) | → | c(a(a(x1))) | (10) |
b(c(c(x1))) | → | a(b(c(x1))) | (11) |
b#(c(c(x1))) | → | a#(b(c(x1))) | (13) |
a#(b(b(x1))) | → | b#(b(b(x1))) | (17) |
The dependency pairs are split into 0 components.