Using the addition function over the natural numbers, give a recursive definition of multiplication of natural numbers?
By : user3081337
Date : March 29 2020, 07:55 AM
it fixes the issue You can think of 3 * 5 as 5 + 5 + 5, i.e. adding 5 for 3 times. If you want to do it recursively, then you can think of it like this: the result of a * b is equal to adding b to the result of (a1) * b. From here to a Haskell recursive function, the step is small :)

Why require a smallest set in a recursive definition of natural numbers?
By : Bharathi Magesh
Date : March 29 2020, 07:55 AM
this one helps. Rule 2 on the list is an "if", not an "if and only if". They're not rules for generating the set, they're rules for deciding if a set is allowed. The set {1, 1.649, 2, 2.649, 3, 3.649, ...} satisfies rule 1, because 1 is in the set. It satisfies rule 2, because for every element of the set, that element plus one is also in the set. In fact, even the set of real numbers satisfies the first two rules, and it has uncountably many "extra" elements that you don't need. Only rule 3 stops you adding arbitrary extra elements to the set, by saying the set has to be the smallest possible one.

What are the steps that Prolog follows with the definition of natural numbers?
By : Vadim Korobka
Date : March 29 2020, 07:55 AM
This might help you It helps to understand how Prolog operates when figuring out an existing predicate, or when designing a new one. When you make a query such as: code :
sum( succ(succ(0)), succ(succ(succ(0))), Answer ).
(1) sum( succ(X), Y, succ(Z) ) : sum( X, Y, Z ).
(2) sum( 0, X, X ).
X = succ(0)
Y = succ(succ(succ(0)))
(A) Answer = succ(Z)
sum( succ(0), succ(succ(succ(0))), Z )
  
X Y Z
X = 0
Y = succ(succ(succ(0)))
(B) Z = succ(Z')
sum( 0, succ(succ(succ(0))), Z' )
  
X Y Z'
0 = 0
X = succ(succ(succ(0)))
(C) X = Z'
Z = succ(Z') = succ(succ(succ(succ(0))))
Answer = succ(Z) = succ(succ(succ(succ(succ(0)))))

Replace integers with consecutive natural numbers in two columns
By : user3209074
Date : March 29 2020, 07:55 AM
Does that help In R, you could convert the desired columns ("A_1", "A_2") to "factor" class by specifying the "levels" as the sorted concatenated unique elements from those columns ("lvls"), then coerce it back to "numeric". code :
lvls < sort(unique(c(D16S503$A_1,D16S503$A_2)))
D16S503[4:5] < lapply(D16S503[4:5], function(x)
as.numeric(factor(x, levels=lvls)))
head(D16S503,3)
# MARKER LANE ID A_1 A_2
#1 D16S503 D16S503 12280198898 4 6
#2 D16S503 D16S503 12280198899 1 5
#3 D16S503 D16S503 12280198900 1 5

Settheoretic definition of natural numbers (recursion)
By : user3582426
Date : March 29 2020, 07:55 AM
like below fixes the issue Just counting open braces won't distinguish wellformed inputs from malformed inputs. Start by writing a program that can construct and output a bracestring form number given an input value. code :
class Set {
std::set<Set> elements;
public:
void read(std::istream in) {...} // recursively reads & adds Sets
int size() {...}
}

