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Proof step 3 from proof primes-divide (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 4:
i < i0 * Π ints0, 0 < i, 0 < i0 * Π ints0, 1 < i0 * Π ints0, ∀ i1. (∀ j. j ∈ ints0 → prime(j) ∧ j | i1) ∧ 0 < i1 → Π ints0 ≤ i1, i0 = i0 ∨ i0 ∈ ints0 → prime(i0) ∧ i0 | i, ∀ j. j ∈ ints0 → prime(j) ∧ j | i ⊦ dups(ints0), i0 ∈ ints0

Rule argument

pos: left 6 terms: i0 abort: F discard: F computed: T

Additional information

Used terms: i0