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Proof step 2 from proof divides-sum (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ i. (∀ j. j ∈ ints0 → i | j) → i | Σ ints0, ∀ j. j = i0 ∨ j ∈ ints0 → i | j ⊦ i | i0 + Σ ints0

Rule argument

Additional information

Used simplifier rules: in-17: ⊦ i ∈ i0 ' ↔ i = i0; from specification list[intlista]%in-append: ⊦ i ∈ ints + ints0 ↔ ¬ ¬(i ∈ ints ∨ i ∈ ints0); from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%sum-rec: ⊦ Σ(i ' + ints) = i + Σ ints; from specification intlist%c: ⊦ i + ints = i ' + ints; from specification list[intlista]