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Proof step 7 from proof max_exp-fac-gen (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
j2 < i, 1 < i ^ i0, 0 ≤ i0, 0 < i ^ i0, i1 * i + j2 < i ^ i0, prime(i), 0 ≤ j2, 0 < i, 0 < i1 * i + j2, ∀ j0, i2, i0. i→n(j0) < i→n(i1 * 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) ⊦ max_exp(i, fac(i1 * i + j2)) = Σ countFun(1, i0, λ j0. (i1 * i + j2) / i ^ j0)

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs