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Proof step 2 from proof prod-eq-zero (previous step)

Π nats0 = 0 ↔ 0 ∈ nats0 ⊦ Π(m + nats0) = 0 ↔ 0 ∈ m + nats0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
Π nats0 = 0 ↔ 0 ∈ nats0 ⊦ m * Π nats0 = 0 ↔ 0 = m ∨ 0 ∈ nats0

Rule argument

Additional information

Used simplifier rules: prod-rec: ⊦ Π(n ' + nats) = n * Π nats;% ⊦ m * 0 = 0;% ⊦ ¬ m ∈ nats → (m ∈ m0 ' + nats ↔ m = m0);% ⊦ m ∈ nats → m ∈ nats0 + nats;% ⊦ ¬ m ∈ m0 ' + nats ↔ m ≠ m0 ∧ ¬ m ∈ nats;% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0;% ⊦ (nats + nats0) + nats1 = nats + nats0 + nats1;% ⊦ m ∈ m0 ' ↔ m = m0;% ⊦ m ∈ nats + nats0 ↔ ¬ ¬(m ∈ nats ∨ m ∈ nats0);% ⊦ m + nats = m ' + nats;