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Proof step 2 from proof potex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ n = 0, ∃ m. 2^ m ≤ n ∧ n < 2 * 2^ m

Rule argument

Additional information

Used simplifier rules: lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ n;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic