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Proof step 11 from proof max_exp-fac-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: max_exp-zero: ⊦ prime(i) ∧ 0 < j0 ∧ j0 < i → max_exp(i, i + j0) = 0;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%zero-02: ⊦ 0 < i ∧ 0 < j → 0 < i + j; from specification int-basic2%ax-04: ⊦ i + 0 = i; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2