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Proof step 4 from proof prime-ex-max (previous step)

1 < i, 2 ≤ i, 0 ≤ i, 0 < i ⊦ ∀ j. prime(j) ∧ j * j ≤ i → ¬ j | i, ∃ j. prime(j) ∧ j | i ∧ j < i ∧ (∀ j0. j < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
j | i, prime(j), j * j ≤ i, 2 ≤ i, 0 ≤ i, 0 < i ⊦ ∃ j. prime(j) ∧ j | i ∧ j < i ∧ (∀ j0. j < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i)

Rule argument

Additional information

Used simplifier rules: numlsledel-01: i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j; from specification int-abs