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