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Proof step 3 from proof product-prime-divides (previous step)

∀ i. prime(i) → (i | Π ints0 ↔ (∃ j. j ∈ ints0 ∧ i | j)), prime(i) ⊦ i | Π(i0 + ints0) ↔ (∃ j. j ∈ i0 + ints0 ∧ i | j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
prime(i), ∀ i. prime(i) → (i | Π ints0 ↔ (∃ j. j ∈ ints0 ∧ i | j)) ⊦ i | i0 * Π ints0 ↔ (∃ j. (j = i0 ∨ j ∈ ints0) ∧ i | j)

Rule argument

Additional information

Used simplifier rules: product-app: ⊦ Π(ints + ints0) = Π ints * Π ints0;%product-one: ⊦ Π i ' = i;%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]%c: ⊦ i + ints = i ' + ints; from specification list[intlista]