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Proof step 9 from proof divides-prime (previous step)

∃ i0. i * i0 = j, 2 ≤ j, 0 ≤ j, 0 < j, ∀ i. 1 < i ∧ i < j → ¬ i | j ⊦ abs(i) = j, abs(i) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
2 ≤ i * i0, 0 ≤ i * i0, 0 < i * i0, ∀ i1. 1 < i1 ∧ i1 < i * i0 → ¬ i1 | i * i0 ⊦ abs(i) = i * i0, abs(i) = 1

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