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Proof step 4 from proof lls-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
2 ^ m0 ≤ m, n < 2 ^ (m0 + 1), 2 ^ m0 ≤ n, log2 m < log2 n, n ≤ m ⊦ 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;%ler: ⊦ m ≤ m; 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%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1