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Proof step 14 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 rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ m0 * m * n = (m0 * m) * n
Proof step 16:
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

Rule argument

spec: nat-mult inst: name: mult-a seq: ⊦ m0 * m * n = (m0 * m) * n subst: [n, m, m0] → [2^ m, 2, 2] flags: rewrite

Additional information

Used terms: 2^ m, 2, 2