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Proof step 5 from proof max_exp-fac-01 (previous step)

0 < i, 0 ≤ j0, j0 < i, prime(i), ∀ 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, fac(i + j0)) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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, fac(i + j0)) = 1

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs