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Proof step 16 from proof potex (previous step)

4 = 2 * 2, 2 * 2^ m = n0 + 1, 2^ m ≤ n0, 2 * 2 * 2^ m ≤ n0 + 1 ⊦ n0 = 0, ∃ m0. 2^ m0 ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^ m0

User applied rule insert equation

And obtained the following new goals

Proof step 17:
4 = 2 * 2, 2 * 2^ m = n0 + 1, 2^ m ≤ n0, 2 * (n0 + 1) ≤ n0 + 1 ⊦ n0 = 0, ∃ m0. 2^ m0 ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^ m0

Rule argument

pos: left 2 flags:

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