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Proof step 6 from proof plus (previous step)

2 * m = 2 * m0 + n ⊦ 2 * (m - m0) = n

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m, m0, n0] flags:

Additional information

Used simplifier rules: dist: ⊦ m0 * (m + n) = m0 * m + m0 * n; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%two-succ-06: ⊦ 2 * m = 2 * n + m0 → ¬ m < n; from specification nat-mult%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m, m0, n0