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Proof step 5 from proof exinpot (previous step)

1 < i, 0 < i, 0 < j, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 → (∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ j0 ∧ j0 < i ^ (i0 + 1)) ⊦ ∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ j ∧ j < i ^ i0 * i, 1 < i ∧ 0 < j - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
1 < i, 0 < i, ∀ j0, i. i→n(j0) < i→n(1) → 1 < i ∧ 0 < j0 → (∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ j0 ∧ j0 < i ^ (i0 + 1)) ⊦ ∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ 1 ∧ 1 < i ^ i0 * i

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i < j - i0 ↔ i + i0 < j; (s)