For each natural number n>0, we define the set B_n to be {2^i:i in ℕ and 0<=i<n}

My inductive definition for B_n is

∀n∈ℕ+B_n={2^0,2^1,2^2,…,2^(n-1)}={B_(n-1)}∪{2^(n-1)}

My proof that for any natural number n>0 if t∈ℕ and 0<=t<2^n , then there is a subset A of B_n such that ∑[x∈A]x=t.

[Base Case]:

Prove P(1)

Let n=1

Then t∈ℕ s.t 0≤t<2^1

So t∈{0,1}

∅⊂B_1 ∑[x∈∅]x = 0

{2^0}⊂B_1 ∑[x∈{2^0}]x = 1

∴∀t∈ℕ s.t 0≤t<2^1 ∃ A⊆B s.t ∑[x∈A]x = t

[I.S]:

Let n≥2

Suppose P(n) [I.H]

W.T.P P(n+1) i.e. t∈ℕ s.t 0≤t<2^(n+1) → ∃ subset A s.t A⊆B_(n+1) ˄ ∑[x∈A]x = t

t∈ℕ s.t 0≤t≤2^(n)-1 → ∃ subset A s.t A⊆B_n ˄ ∑[x∈A]x = t [re-arranging inequality]

For n

C1:

0≤t≤2^(n-1)-1 [since t∈ℕ, n≥2] (*) [re-arranging inequality]

C2:

2^(n-1)-1<t≤2^(n)-1 [since t∈ℕ, n≥2] (**) [re-arranging inequality]

For n+1

C1:

0≤t≤2^(n-1+1)-1 [since t∈ℕ, n≥2] (***) [by I.H, since ***=*+**]

C2:

2^(n-1+1)-1<t≤2^(n+1)-1 [since t∈ℕ, n≥2] [by I.H]

0≤t<2n+1 as wanted.

Is this right and formal?