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Proof step 11 from proof lls-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m < 2 ^ (m1 + 1), log2 m < m0, n < 2 ^ (m1 + 1), n < 2 ^ (m0 + 1), log2 m < m1, 2 ^ m1 ≤ m, n ≤ m, 2 ^ m0 ≤ n, 2 ^ m0 ≤ m, m0 ≤ m1 ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: pot2-def: ⊦ 2^ m = 2 ^ m;% ⊦ 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% ⊦ m +1 = m + 1; (s)