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Proof step 4 from proof twodiv-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: one-03: ⊦ 1 = m + 1 ↔ m = 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1