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Proof step 13 from proof max_exp-fac-add (previous step)

0 < fac(i * j) ∧ 0 < Π countFun(1, j0, λ j0. i * j + j0) ∧ prime(i), prime(i), j0 < i, 0 < j, 0 ≤ j0, 0 < i, 0 ≤ j ⊦ max_exp(i, fac(i * j)) + max_exp(i, Π countFun(1, j0, λ j0. i * j + j0)) = max_exp(i, fac(i * j))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: 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%eq: ⊦ i = i + j ↔ j = 0; from specification int-basic2