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Proof step 8 from proof max_exp-fac-smaller (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
1 < i, 0 < i, 0 < j, j < i, prime(i), i | fac(j - 1), 0 < fac(j - 1), ∀ j0, i. i→n(j0) < i→n(j) → prime(i) ∧ j0 < i ∧ 0 < j0 ∧ 0 < i ∧ 1 < i → max_exp(i, fac(j0)) = 0 ⊦

Rule argument

Additional information

Used simplifier rules: max_exp-smaller: ⊦ prime(i) ∧ 0 < j ∧ j < i → max_exp(i, j) = 0;%max_exp-zero-02: ⊦ prime(i) ∧ 0 < i0 → (max_exp(i, i0) = 0 ↔ ¬ i | i0);%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) 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%ax-04: ⊦ i + 0 = i; from specification int-basic2