The rewrite relation of the following TRS is considered.
a(a(x1)) | → | b(b(x1)) | (1) |
c(c(b(x1))) | → | d(c(a(x1))) | (2) |
a(x1) | → | d(c(c(x1))) | (3) |
c(d(x1)) | → | b(c(x1)) | (4) |
a(a(x1)) | → | b(b(x1)) | (1) |
b(c(c(x1))) | → | a(c(d(x1))) | (5) |
a(x1) | → | c(c(d(x1))) | (6) |
d(c(x1)) | → | c(b(x1)) | (7) |
a#(a(x1)) | → | b#(b(x1)) | (8) |
a#(a(x1)) | → | b#(x1) | (9) |
b#(c(c(x1))) | → | a#(c(d(x1))) | (10) |
b#(c(c(x1))) | → | d#(x1) | (11) |
a#(x1) | → | d#(x1) | (12) |
d#(c(x1)) | → | b#(x1) | (13) |
The dependency pairs are split into 1 component.
a#(x1) | → | d#(x1) | (12) |
d#(c(x1)) | → | b#(x1) | (13) |
b#(c(c(x1))) | → | a#(c(d(x1))) | (10) |
b#(c(c(x1))) | → | d#(x1) | (11) |
[d(x1)] | = | 1 · x1 |
[c(x1)] | = | 1 · x1 |
[b(x1)] | = | 1 · x1 |
[a(x1)] | = | 1 · x1 |
[d#(x1)] | = | 1 · x1 |
[a#(x1)] | = | 1 · x1 |
[b#(x1)] | = | 1 · x1 |
d(c(x1)) | → | c(b(x1)) | (7) |
b(c(c(x1))) | → | a(c(d(x1))) | (5) |
a(x1) | → | c(c(d(x1))) | (6) |
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Hence, it suffices to show innermost termination in the following.[a#(x1)] | = |
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[d#(x1)] | = |
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[c(x1)] | = |
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[b#(x1)] | = |
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[d(x1)] | = |
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[b(x1)] | = |
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[a(x1)] | = |
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b#(c(c(x1))) | → | d#(x1) | (11) |
[a#(x1)] | = |
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[d#(x1)] | = |
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[c(x1)] | = |
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[b#(x1)] | = |
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[d(x1)] | = |
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[b(x1)] | = |
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[a(x1)] | = |
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a#(x1) | → | d#(x1) | (12) |
The dependency pairs are split into 0 components.