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Proof step 4 from proof pot-square-est-01 (previous step)

4 ^ i ≤ i * i, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 4 ^ i0 ≤ i0 * i0 ∧ 0 ≤ i0 → false ⊦

User applied rule apply induction

And obtained the following new goals

Proof step 5:
4 ^ i ≤ i * i, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 4 ^ i0 ≤ i0 * i0 ∧ 0 ≤ i0 → false ⊦ 4 ^ (i - 1) ≤ (i - 1) * (i - 1) ∧ 0 ≤ i - 1

Rule argument

subst: [i0] → [i - 1]

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%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: i - 1