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Proof step 9 from proof potex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: zero: ⊦ 2^ n ≠ 0;%ref-1: ⊦ ¬ n < n; from specification nat-basic1%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lels-03: ⊦ m < n → m ≤ n; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%les: ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%ls-07: ⊦ n0 < n → (n < m * n ↔ 1 < m); from specification nat-mult%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat