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Proof step 24 from proof max_exp-fac-rec (previous step)

0 < i, 0 ≤ j0, j0 < i, prime(i), 0 < j, 0 < fac(i * (j - 1) + (i - 1)), 0 < fac(j - 1), ∀ j1, i, j0. i→n(j1) < i→n(j) → true → prime(i) ∧ 0 < j1 ∧ 0 ≤ j0 ∧ j0 < i → max_exp(i, fac(i * j1 + j0)) = j1 + max_exp(i, fac(j1)) ⊦ prime(i) ∧ 0 < j - 1 ∧ 0 ≤ i - 1 ∧ i - 1 < i, 1 + max_exp(i, fac(i * (j - 1) + (i - 1))) = j + max_exp(i, fac(j - 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: fac-zero: ⊦ i ≤ 0 → fac(i) = 1;%max_exp-one: ⊦ prime(i) → max_exp(i, 1) = 0;%max_exp-zero-02: ⊦ prime(i) ∧ 0 < i0 → (max_exp(i, i0) = 0 ↔ ¬ i | i0);%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%ax-04: ⊦ i + 0 = i; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%Mult-zero: ⊦ i * 0 = 0; from specification int-mult%sub-eq-01: ⊦ i = j → i - j = 0; from specification int-abs%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%ler: ⊦ i ≤ i; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < 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-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs