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Proof step 10 from proof lls-02 (previous step)

2^ m0 ≤ n ∧ n < 2^ m0 +1, m < 2 ^ (m1 + 1), 2 ^ m1 ≤ m, n ≤ m, log2 m < m0, 2 ^ m0 ≤ n, n < 2 ^ (m0 + 1), 2 ^ m0 ≤ m, n < 2 ^ (m1 + 1), m0 ≤ m1 ⊦ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
m < 2 ^ (m1 + 1), 2 ^ m1 ≤ m, n ≤ m, log2 m < m0, 2 ^ m0 ≤ m, n < 2 ^ (m1 + 1), m0 ≤ m1, 2 ^ m0 ≤ n, n < 2 ^ (m0 + 1), log2 m < m1 ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: pot2-def: ⊦ 2^ m = 2 ^ m;%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ: ⊦ m +1 = m + 1; from specification nat-basic2