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Proof step 3 from proof prod-eq-one-01 (previous step)

∀ m. Π nats0 = 1 ∧ m ∈ nats0 → m = 1 ⊦ Π(m + nats0) = 1 ∧ m0 ∈ m + nats0 → m0 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
Π nats0 = 1, m0 ∈ nats0, ∀ m1. m1 ∈ nats0 → m1 = 1 ⊦ m0 = 1

Rule argument

Additional information

Used simplifier rules: prod-rec: ⊦ Π(n ' + nats) = n * Π nats;% ⊦ m ∈ m0 ' ↔ m = m0;% ⊦ m ∈ nats + nats0 ↔ ¬ ¬(m ∈ nats ∨ m ∈ nats0);% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ m0 * m = 1 ↔ m0 = 1 ∧ m = 1;% ⊦ (nats + nats0) + nats1 = nats + nats0 + nats1;% ⊦ m + nats = m ' + nats;