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Proof step 51 from proof max_exp-fac-gen (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-basic1 inst: name: ls-rec seq: ⊦ i ≠ j -1 → (i < j ↔ i < j -1) subst: [j, i] → [n→i(n) + 2, i0] flags: rewrite

Additional information

Used simplifier rules: sub-pred: ⊦ i -1 = i - 1; from specification int-abs%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n; from specification intnat%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%ls-int-ls-nat: ⊦ i ≥ 0 → ((* i) + n→i(n) < n→i(m) ↔ i→n(i) + n < m); from specification intnat%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%ax-04: ⊦ i + 0 = i; from specification int-basic2%nat2int-07: ⊦ 0 ≠ n→i(m) + 1; from specification intnat%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2 Used terms: n→i(n) + 2, i0