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Proof step 14 from proof potex (previous step)

2 * 2^ m = n0 + 1, 4 = 2 * 2, 4 * 2^ m ≤ n0 + 1, 2^ m ≤ n0, 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 15:
2 * 2^ m = n0 + 1, 4 = 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

Rule argument

pos: left 2 flags:

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