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Proof step 4 from proof divides-sum (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%divides-add: ⊦ i | j → (i | j + j0 ↔ i | j0); from specification int-gcd