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Proof step 17 from proof potex (previous step)

4 = 2 * 2, 2 * 2^ m = n0 + 1, 2^ m ≤ n0, 2 * (n0 + 1) ≤ n0 + 1 ⊦ n0 = 0, ∃ m0. 2^ m0 ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^ m0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ls-01: ⊦ m0 * m < m ↔ m0 = 0 ∧ m ≠ 0; from specification nat-mult%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%le-num-02: n0 < n ⊦ m + (* n) ≤ m0 + (* n0) ↔ m + (n - n0) ≤ m0; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult