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Proof step 16 from proof max_exp-fac-gen (previous step)

0 ≤ i0, 1 < i ^ i0, j2 < i, 0 < i ^ i0, 0 ≤ j2, prime(i), j2 < i ^ i0, 1 < i, 0 < i, 0 < j2, ∀ j0, i2, i0. i→n(j0) < i→n(0 * i + j2) → prime(i2) ∧ j0 < i2 ^ i0 ∧ 0 < j0 ∧ 0 < i2 ^ i0 ∧ 0 ≤ i0 ∧ 1 < i2 ^ i0 → max_exp(i2, fac(j0)) = Σ countFun(1, i0, λ j1. j0 / i2 ^ j1) ⊦ 0 = Σ countFun(1, i0, λ j. j2 / i ^ j), (0 < i1 ∧ i1 ≤ i0 → j2 / i ^ i1 = 0) ∧ 0 < i0

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

Additional information

Used simplifier rules: lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lel: ⊦ i < i0 ∧ i = i1 → i1 < i0; (forward) from specification oelem[intlista]%Pot-zero: ⊦ j ^ 0 = 1; from specification int-pot%bet-01: ⊦ 0 ≤ i → (i < 1 ↔ i = 0); from specification int-abs