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Proof step 19 from proof prime-def-mod (previous step)

0 ≤ i0, 2 ≤ i * i0, 0 < i, 2 < i * i0, 1 < i, 1 < i0, 0 < i0, 1 < i ∧ i * i ≤ i * i0 → (i * i0) % i ≠ 0, ∀ i1. 1 < i1 ∧ i1 * i1 ≤ i * i0 → (i * i0) % i1 ≠ 0 ⊦ i * i0 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: numlslsdel-01: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j; from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0; from specification int-div%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%zero-left: ⊦ 0 * i = 0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%le-pos: ⊦ 0 < i → (i * j ≤ i * i0 ↔ j ≤ i0); from specification int-mult%numlelsdel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j; from specification int-abs