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Proof step 12 from proof divides-prime (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: div-mult: ⊦ i | i * j;% ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)% ⊦ i * j = i ↔ i = 0 ∨ j = 1; (ws)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ 0 < i → (0 < i * j ↔ 0 < j); (s)% ⊦ 0 < i → (i * j ≤ i ↔ j ≤ 1); (s)%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%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% ⊦ i < j ↔ ¬ j ≤ i; (ss)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult