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Proof step 10 from proof primes-divide (previous step)

i < i0 * Π ints0, 0 < i, 0 < i0 * Π ints0, 1 < i0 * Π ints0, prime(i0), ∃ i1. i0 * i1 = i, j ∈ ints0 → prime(j) ∧ j | i, ∀ j. j ∈ ints0 → prime(j) ∧ j | i, (j ∈ ints0 → prime(j) ∧ j | i / i0) ∧ 0 < i / i0 → Π ints0 ≤ i / i0 ⊦ dups(ints0), i0 ∈ ints0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i0 * i1 < i0 * Π ints0, 0 < i0 * Π ints0, 1 < i0 * Π ints0, prime(i0), 0 < i0 * i1, j ∈ ints0 → prime(j) ∧ j | i0 * i1, ∀ j. j ∈ ints0 → prime(j) ∧ j | i0 * i1, (j ∈ ints0 → prime(j) ∧ j | i1) ∧ 0 < i1 → Π ints0 ≤ i1 ⊦ i0 = 0, dups(ints0), i0 ∈ ints0

Rule argument

Additional information

Used simplifier rules: ls-forward: ⊦ i * j < i * j0 → i ≠ 0; (forward) from specification int-mult%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%div-mult-03: j ≠ 0 ⊦ (i * j) / j = i; from specification int-div%prime-not-neg-01: ⊦ j ≤ 0 → ¬ prime(j); from specification int-gcd%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%Mult-zero: ⊦ i * 0 = 0; from specification int-mult