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Proof step 10 from proof potex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
2 * 2^ m = n0 + 1, 2^ m ≤ n0, 2^ m ≤ n0 + 1 ⊦ n0 = 0, ∃ m0. 2^ m0 ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^ m0

Rule argument

Additional information

Used simplifier rules: zero: ⊦ 2^ n ≠ 0;% ⊦ ¬ n < n; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n → m ≤ n; (s)% ⊦ m + 1 ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%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% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)% ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ n0 < n → (n < m * n ↔ 1 < m); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)