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Proof step 3 from proof exinpot (previous step)

0 < j, 0 < i, 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 * i

User applied rule apply induction

And obtained the following new goals

Proof step 4:
0 < j, 0 < i, 1 < i, ∀ j0, i. i→n(j0) < i→n(j) → 0 < j0 ∧ 1 < i → (∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ j0 ∧ j0 < i ^ (i0 + 1)) ⊦ 0 < j - 1 ∧ 1 < i, ∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ j ∧ j < i ^ i0 * i
Proof step 6:
∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ j - 1 ∧ j - 1 < i ^ (i0 + 1), 0 < j, 0 < i, 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 * i

Rule argument

subst: [j0, i] → [j - 1, i]

Additional information

Used simplifier rules: abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: j - 1, i