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Proof step 44 from proof max_exp-fac-gen (previous step)

0 ≤ 2 ∧ 2 ≤ i0, max_exp(i, fac(i1)) = Σ countFun(1, i0 - 1, λ j1. i1 / i ^ j1), (i1 * i + j2) / i = i1, 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, 1 < i0, 0 < i0, ∀ 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(i0) - 1 = i→n(1 + i0 - 2) ∧ (n + 1 < i→n(i0) → 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: numlsledel-01: i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%abs-zero: ⊦ i→n(i) = 0 ↔ i = 0; from specification intnat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%pred-03: ⊦ n -1 = n - 1; from specification nat%rax-01: ⊦ 1 < i → i→n(i - 1) = i→n(i) -1; from specification intnat%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs