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Proof step 4 from proof exinpot (previous step)

0 < j, 0 < i, 1 < i, ∀ 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

User applied rule apply induction

And obtained the following new goals

Proof step 5:
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
Proof step 7:
1 < i, 0 < i, 0 < j, ∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ j - 1 ∧ j - 1 < i ^ (i0 + 1), ∀ 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

Rule argument

subst: [j0, i] → [j - 1, i]

Additional information

Used simplifier rules: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: j - 1, i