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Proof step 3 from proof exin2pot (previous step)

0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ j0 ∧ j0 < 2 ^ (i + 1)) ⊦ ∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1)

User applied rule cut formula

And obtained the following new goals

Proof step 4:
0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ j0 ∧ j0 < 2 ^ (i + 1)) ⊦ j = 0, ∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1)

Rule argument

fma: j = 0

Additional information

Used formulas: j = 0