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Proof step 7 from proof modeq (previous step)

(m1 * n + n) + n0 = m0 * n + n1, n1 < n, n0 < n ⊦ m1 * n + n0 = (m0 - 1) * n + n1, n = 0, m1 + 1 = m0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m0, m] flags:

Additional information

Used simplifier rules: zero: ⊦ 0 * n = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%rsucc-01: ⊦ (m + 1) * n = m * n + n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%l-05: ⊦ m + m0 < n → ¬ n + n0 < m; from specification nat-basic2%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m0, m