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Proof step 2 from proof two-loup-10 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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 + n, 2, m0, n0] flags: rewrite, current

Additional information

Used simplifier rules: numeral-19: m2 ≠ 0 ⊦ (* m2) * m + n = m + n0 ↔ (m2 - 1) * m + n = n0; from specification nat-mult%one: ⊦ m0 * 1 = m0; from specification nat-mult%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-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%twols-02: ⊦ m ≠ 0 → (m < 2 ↔ m = 1); from specification nat-basic1 Used terms: m + n, 2, m0, n0