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Proof step 22 from proof divides-prime (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 5

Additional information

Used simplifier rules: neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%zero-gr-01: ⊦ i < 0 → (0 < i * j ↔ j < 0); from specification int-mult%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%mone-08: ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); from specification int-abs%neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs