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Proof step 4 from proof twodiv-02 (previous step)

1 < m ∧ (∀ m0. m0 | m → m0 = 1 ∨ m0 = m) ⊦ 2 | m + 1 ↔ m ≠ 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ m0. m0 | m → m0 = 1 ∨ m0 = m ⊦ m = 1, m = 0, m = 2, 2 | m + 1

Rule argument

Additional information

Used simplifier rules: divi-three: ⊦ m ≠ 1 → (m | 3 ↔ m = 3);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1