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Proof step 2 from proof potex (previous step)

⊦ n = 0, ∃ m. 2^ m ≤ n ∧ n < 2 * 2^ m

User applied rule structural induction

And obtained the following new goals

Proof step 3:
n0 = 0 ∨ (∃ m. 2^ m ≤ n0 ∧ n0 < 2 * 2^ m) ⊦ n0 +1 = 0, ∃ m. 2^ m ≤ n0 +1 ∧ n0 +1 < 2 * 2^ m

Rule argument

indvar: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;