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Proof step 15 from proof dec-nat-01 (previous step)

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

User applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ (2 * m + n) / 2 = m + n / 2; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≠ 0 → (m < 2 ↔ m = 1); (s)% ⊦ m * (n + 1) = m * n + m; (s)%n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s)%n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: n, 2, m0, n0