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Proof step 5 from proof lls (previous step)

n < 2 ^ log2 n, ∃ m. 2^ m ≤ n ∧ n < 2^(m + 1) ⊦ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m < log2 n, n < 2 ^ log2 n, n < 2 ^ (m + 1), 2 ^ m ≤ n ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: pot2-def: ⊦ 2^ m = 2 ^ m;%two-pot-zero: ⊦ 2 ^ n ≠ 0;%pot-mon: 1 < m ⊦ m ^ n < m ^ n0 ↔ n < n0;% ⊦ m < n → n ≠ 0; (s)% ⊦ m ≤ m; (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 + 1 ↔ ¬ n < m; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic