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Proof step 3 from proof potex (previous step)

n0 = 0 ∨ (∃ m. 2^ m ≤ n0 ∧ n0 < 2 * 2^ m) ⊦ n0 +1 = 0, ∃ m. 2^ m ≤ n0 +1 ∧ n0 +1 < 2 * 2^ m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 = 0 ∨ 2^ m ≤ n0 ∧ n0 < 2 * 2^ m ⊦ ∃ m0. 2^ m0 ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^ m0

Rule argument

Additional information

Used simplifier rules: 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%succ: ⊦ m +1 = m + 1; from specification nat-basic2