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Proof step 11 from proof potex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
2 * 2^ m = n0 + 1, 2^ m ≤ n0, 4 * 2^ m ≤ n0 + 1 ⊦ n0 = 0, ∃ m0. 2^ m0 ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^ m0

Rule argument

Additional information

Used simplifier rules: lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ n;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2