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Proof step 5 from proof potex (previous step)

n0 = 0, n0 = 0 ∨ 2^ m ≤ n0 ∧ n0 < 2 * 2^ m ⊦ ∃ m0. 2^ m0 ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^ m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ ∃ m. 2^ m ≤ 1

Rule argument

Additional information

Used simplifier rules: zero: ⊦ 2^ n ≠ 0;%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%ls2-03: ⊦ 1 < 2 * n ↔ n ≠ 0; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2