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Proof step 14 from proof dec-int (previous step)

2 * i→n(i) = m + 1, i < 0 ⊦ i = 0, m % 2 = 0, n→i((m + 1) / 2) + i = 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
m = m0 * 2 + n0 ∧ n0 < 2 ∧ 2 ≠ 0, 2 * i→n(i) = m + 1, i < 0 ⊦ i = 0, n0 = 0, n→i((m + 1) / 2) + i = 0

Rule argument

spec: nat-div inst: name: div-elim seq: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n subst: [m, n, m0, n0] → [m, 2, m0, n] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: m, 2, m0, n