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Proof step 1 from proof exinpot (conclusion)

0 < j, 1 < i ⊦ ∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ j ∧ j < i ^ (i0 + 1)

User applied rule induction

And obtained the following new goals

Proof step 2:
0 < j ∧ 1 < i, ∀ j0, i. i→n(j0) < i→n(j) → 0 < j0 ∧ 1 < i → (∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ j0 ∧ j0 < i ^ (i0 + 1)) ⊦ ∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ j ∧ j < i ^ (i0 + 1)

Rule argument

indvar: i→n(j) wfpred: <

Additional information

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