The rewrite relation of the following TRS is considered.
a(x1) | → | x1 | (1) |
a(a(b(x1))) | → | c(c(b(a(x1)))) | (2) |
b(c(x1)) | → | a(b(x1)) | (3) |
a(x1) | → | x1 | (1) |
b(a(a(x1))) | → | a(b(c(c(x1)))) | (4) |
c(b(x1)) | → | b(a(x1)) | (5) |
b#(a(a(x1))) | → | c#(x1) | (6) |
b#(a(a(x1))) | → | c#(c(x1)) | (7) |
b#(a(a(x1))) | → | b#(c(c(x1))) | (8) |
b#(a(a(x1))) | → | a#(b(c(c(x1)))) | (9) |
c#(b(x1)) | → | a#(x1) | (10) |
c#(b(x1)) | → | b#(a(x1)) | (11) |
The dependency pairs are split into 1 component.
c#(b(x1)) | → | b#(a(x1)) | (11) |
b#(a(a(x1))) | → | c#(x1) | (6) |
b#(a(a(x1))) | → | c#(c(x1)) | (7) |
b#(a(a(x1))) | → | b#(c(c(x1))) | (8) |
π(c#) | = | { 1, 1 } |
π(b#) | = | { 1, 1 } |
π(c) | = | { 1 } |
π(b) | = | { 1, 1 } |
π(a) | = | { 1 } |
c#(b(x1)) | → | b#(a(x1)) | (11) |
The dependency pairs are split into 1 component.
b#(a(a(x1))) | → | b#(c(c(x1))) | (8) |
[b#(x1)] | = |
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[b(x1)] | = |
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[a(x1)] | = |
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[c(x1)] | = |
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a(x1) | → | x1 | (1) |
b(a(a(x1))) | → | a(b(c(c(x1)))) | (4) |
c(b(x1)) | → | b(a(x1)) | (5) |
b#(a(a(x1))) | → | b#(c(c(x1))) | (8) |
There are no pairs anymore.