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Proof step 6 from proof divides-prime (previous step)

i | j, 2 ≤ 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 7:
i | j, 2 ≤ j, 0 ≤ j, 0 < j, ∀ i. 1 < i ∧ i < j → ¬ i | j ⊦ abs(i) = j, 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: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs