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Proof step 7 from proof exin2pot (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: 0

Additional information

Used terms: 0