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Proof step 2 from proof exin2pot (previous step)

0 < j ⊦ ∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1)

User applied rule induction

And obtained the following new goals

Proof step 3:
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)

Rule argument

indvar: i→n(j) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j)