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Proof step 5 from proof divides-sum (previous step)

i | i0, ∀ j. j ∈ ints0 → i | j, ∀ i. (∀ j. j ∈ ints0 → i | j) → i | Σ ints0 ⊦ i | Σ ints0

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: i abort: F discard: T computed: T

Additional information

Used terms: i