Question

Give me a evidence/proof that how '1 + 1 = 2'? Sounds easy, Right?? But very hard to proof this.
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Answer 1

This is hard but hope u can understand ...

Let N be the entire set of natural numbers {1, 2, 3, 4....}, let count x be 1 then 2 then 3 then 4 then...then x. Let n be a list of natural numbers. Let H(n) be {x1} if n = {x1, x2, x3...xn} or, the head of the element in the list, Let T(n) be {x2, x3, x4...xn} or the tail of the list. Let C(x) be T(N) recursively count x-1, or T(N) = {1,2,3,4...} once, T({1,2,3,4...} = {2,3,4,5...} twice, T({2,3,4,5...} = {3,4,5,6..} three times...x times.

Now let x + y be: C(x) = n, count y of H(T(n)) recursively i.e. H(T(n)) becomes n in the subsequent iterations.

So 1 + 1: C(1), count 1 of H (T(n)). C(1) means count N to 1, then you are left with the tail of N once so (1,2,3,4,5...}, now this is set to n (just a list of natural numbers. Now take this n and plug it into H(T(n) which gives you first T{1,2,3...} = {2,3,4...} and then H{2,3,4...} = {2}.

So by definition of 1 + 1, we get {2} or 2.

The idea is you can define x + y where x is the starting point of the list of natural numbers so if x is 5, then start the list with 5 i.e. {5,6,7,8...}. y is the number of iterations you want to take 1 step "forward" so if y is 3, then you start with 5, take 1 step forward 3 times gives you 8.