我在这里做错了什么（我没有足够的经验来完成证明，还是之前有一些错误）？我无法完全证明这两种类型之间的等效性。任何帮助表示赞赏。
这是我为证明log_2（n！）进行排序所必需的一部分（特定于n = 8）。 var data = [{
values: [],
labels: ['Bank Accounts', 'Savings Accounts', 'Cash-on-Hand'],
type: 'pie'
}];
var layout = {
height: 400,
width: 500
};
function testPER() {
var netSalaryPER =
Number($("input#netSalary").val()) /
Number($("input#totalIncomeValue").val())
* 100;
var bonusSalaryPER =
Number($("input#bonusSalary").val()) /
Number($("input#totalIncomeValue").val())
* 100;
var gigSalaryPER =
Number($("input#gigSalary").val()) /
Number($("input#totalIncomeValue").val())
* 100;
data[0].values[0] = netSalaryPER;
data[0].values[1] = bonusSalaryPER
data[0].values[2] = gigSalaryPER
/* TESTING PURPOSES $("input#netSalaryPER").val(netSalaryPER); */
Plotly.newPlot('incomePieChart', data, layout);
}
类型对应于决策树。
algorithm''答案 0 :(得分：3)
您的陈述似乎是错误的。这是一个反例。
Definition mya1 : algorithm' 2 :=
conditional' (GT x0 x1)
((assign (num x0) (num x2)) :: (assign (num x1) (num x3)) :: nil)
((assign (num x0) (num x4)) :: (assign (num x1) (num x5)) :: nil)
(conditional' (GT x2 x3)
((assign (num x2) (num x6)) :: (assign (num x3) (num x7)) :: nil)
((assign (num x0) (num x4)) :: (assign (num x1) (num x5)) :: nil)
leaf' leaf')
(conditional' (GT x3 x0)
((assign (num x2) (num x7)) :: (assign (num x3) (num x6)) :: nil)
((assign (num x0) (num x4)) :: (assign (num x1) (num x5)) :: nil)
leaf' leaf').
Definition myi x :=
if variable_eq_dec x (num x0) then 3 else
if variable_eq_dec x (num x1) then 4 else
if variable_eq_dec x (num x2) then 6 else
if variable_eq_dec x (num x3) then 5 else
if variable_eq_dec x (num x4) then 7 else
if variable_eq_dec x (num x5) then 8 else
if variable_eq_dec x (num x6) then 9 else
0.
以下命令返回6
Compute run_algorithm' mya1 myi (num x2).
以下内容返回0。
Compute run_algorithm'' (algorithm'_to_algorithm'' mya1) myi (num x2).
我通过尝试对a进行归纳证明来找到反例。这是我的脚本。
Theorem algorithm'_algorithm''_equiv n (a: algorithm' n):
forall (i: instantation) (x: variable),
run_algorithm' a i x = run_algorithm'' (algorithm'_to_algorithm'' a) i x.
Proof.
induction a.
easy.
simpl.
intros i x.
destruct c.
destruct (Compare_dec.gt_dec (i (num more)) (i (num less))) as [cmp1 | cmp2].
rewrite IHa1.
这将我引向一个与append_assigments和run_assignment_list有关的目标，我以这种方式陈述了这一观点。
Lemma algorithm''_list n (a : algorithm'' n) (l: list assignment) i x :
run_algorithm'' a (run_assignment_list l i) x =
run_algorithm'' (append_assignments a l) i x.
在做这个证明的过程中，通过对a的归纳，我可以看到在平等的两个方面之间存在执行不匹配的情况。对其进行进一步分析得出了反例。