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Proof step 2 from proof prod-eq-one-01 (previous step)

Π nats = 1, m ∈ nats ⊦ m = 1

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ m. Π nats0 = 1 ∧ m ∈ nats0 → m = 1 ⊦ Π(m + nats0) = 1 ∧ m0 ∈ m + nats0 → m0 = 1

Rule argument

indvar: nats

Additional information

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