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Proof step 2 from proof max_exp-fac-rec (previous step)

true, ∀ 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 ∧ 0 ≤ j0 ∧ j0 < i → max_exp(i, fac(i * j + j0)) = j + max_exp(i, fac(j))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
prime(i), j0 < i, 0 < j, 0 ≤ j0, 0 < i, ∀ 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)) ⊦ max_exp(i, fac(i * j + j0)) = j + max_exp(i, fac(j))

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs