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Proof step 4 from proof max_exp-fac-gen (previous step)

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)

User applied rule cut formula

And obtained the following new goals

Proof step 5:
∃ j0. j0 = j / i, 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)

Rule argument

fma: ∃ j0. j0 = j / i

Additional information

Used formulas: ∃ j0. j0 = j / i