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Proof step 12 from proof potex (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 13:
4 = 2 * 2, 2 * 2^ m = n0 + 1, 2^ m ≤ n0, 4 * 2^ m ≤ n0 + 1 ⊦ n0 = 0, ∃ m0. 2^ m0 ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^ m0

Rule argument

fma: 4 = 2 * 2

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used formulas: 4 = 2 * 2