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Proof step 18 from proof prime-def-mod (previous step)

1 < i, 0 < i, i < j, 0 < j, 0 ≤ j, 2 ≤ j, ∀ i. 1 < i ∧ i * i ≤ j → j % i ≠ 0, ∃ i0. i * i0 = j ⊦ j = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
1 < i, 0 < i, 1 < i0, 0 < i0, 2 ≤ i * i0, 0 ≤ i0, ∀ i1. 1 < i1 ∧ i1 * i1 ≤ i * i0 → (i * i0) % i1 ≠ 0 ⊦ i * i0 = 1

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%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%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1% ⊦ 0 < i → (i < i * j ↔ 1 < j); (s)% ⊦ 0 < i → (0 < i * j ↔ 0 < j); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)