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Proof step 5 from proof primes-divide (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 6:
i < i0 * Π ints0, 0 < i, 0 < i0 * Π ints0, 1 < i0 * Π ints0, prime(i0), i0 | i, ∀ j. j ∈ ints0 → prime(j) ∧ j | i, (∀ j. j ∈ ints0 → prime(j) ∧ j | i / i0) ∧ 0 < i / i0 → Π ints0 ≤ i / i0 ⊦ dups(ints0), i0 ∈ ints0

Rule argument

pos: left 8 terms: i / i0 abort: F discard: T computed: F

Additional information

Used terms: i / i0