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       <dc:date>2026-04-27T19:39:25+00:00</dc:date>
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        <dc:date>2014-10-15T10:59:35+00:00</dc:date>
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        <title>Числа Фиббоначи</title>
        <link>https://wiki.nsunc.com/haskell/fibb?rev=1413370775&amp;do=diff</link>
        <description>Числа Фиббоначи

Рассмотрим лобовое решение с бинарной рекурсией:


fib :: Int -&gt; Int
-- fib :: Integer -&gt; Integer избыточно, так как не достигнем
fib 1 = 1
fib 2 = 1
fib n = fib (n-1) + fib (n-2)</description>
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        <dc:date>2010-10-16T10:03:57+00:00</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title></title>
        <link>https://wiki.nsunc.com/haskell/primex?rev=1287223437&amp;do=diff</link>
        <description>Решение. Разобъем решение на несколько шагов.

1. Введем свой предикат, показывающий, что a делится нацело на b, т.е. a:b (или тоже самое b|a):


de a b = if ((rem a b) == 0) then True else False</description>
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        <dc:date>2013-04-17T11:01:48+00:00</dc:date>
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        <title></title>
        <link>https://wiki.nsunc.com/haskell/sekv?rev=1366196508&amp;do=diff</link>
        <description>data LogicFormula = A|B|C|D|E|F|G|H|I|J|K|L|M|N| 
       LogicFormula :&amp;: LogicFormula | 
       LogicFormula :+: LogicFormula | 
       LogicFormula :-&gt; LogicFormula |     
       Neg LogicFormula deriving (Eq,Read,Show)
data Sekv = [LogicFormula] :- [LogicFormula] deriving (Eq,Read,Show)

aks :: Sekv -&gt; Bool
aks ([x] :- [y]) = if x == y then True else False
ax w = if (aks w) then w else error &quot;w is not an axiom!&quot;

p1  (g :- [x]) (g2 :- [y]) | g==g2     = g :- [x :&amp;: y] 
                       …</description>
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