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Proof step 2 from proof product-negative (previous step)

Π ints < 0, ∀ i. i ∈ ints → 0 ≤ i ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 3:
¬(Π ints0 < 0 ∧ (∀ i. i ∈ ints0 → 0 ≤ i)), Π(i + ints0) < 0, ∀ i0. i0 ∈ i + ints0 → 0 ≤ i0 ⊦

Rule argument

indvar: ints

Additional information

Used simplifier rules: prod-empty: ⊦ Π [] = 1;%intlist freely generated by [], +;;