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Proof step 41 from proof max_exp-fac-gen (previous step)

0 ≤ 1 ∧ 1 ≤ i0 - 1, max_exp(i, fac(i1)) = Σ countFun(1, i0 - 1, λ j1. i1 / i ^ j1), (i1 * i + j2) / i = i1, 0 < i0, 0 < i1, 0 < i1 * i + j2, 0 < i, 0 ≤ j2, prime(i), i1 * i + j2 < i ^ i0, 0 < i ^ i0, 0 ≤ i0, 1 < i ^ i0, j2 < i, ∀ 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)) = Σ countFun(2, i0, λ j0. (i1 * i + j2) / i ^ j0), i→n(1 + (i0 - 1) - 1) = # countFun(2, i0, λ j0. (i1 * i + j2) / i ^ j0) ∧ (n < i→n(1 + (i0 - 1) - 1) → countFun(1, i0 - 1, λ j1. i1 / i ^ j1)[n] = countFun(2, i0, λ j0. (i1 * i + j2) / i ^ j0)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%ls: n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; from specification nat%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%num-i2n-le-02: 0 ≤ i ⊦ (* m) ≤ i→n(i) ↔ n→i(m) ≤ i; from specification intnat%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%pred-03: ⊦ n -1 = n - 1; from specification nat%rax-01: ⊦ 1 < i → i→n(i - 1) = i→n(i) -1; from specification intnat%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs