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Proof step 1 from proof mod-neq (conclusion)

⊦ m0 < m ∧ n0 < n ∧ n1 ≤ n → m0 * n + n0 ≠ m * n + m + n1

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ m0 < m ∧ n ≤ n0 ∧ n2 < n0 ∧ n1 < n → m0 * n + n1 < m * n0 + n2

Rule argument

spec: nat-mult inst: name: mod-ls seq: ⊦ m0 < m ∧ n ≤ n0 ∧ n2 < n0 ∧ n1 < n → m0 * n + n1 < m * n0 + n2 subst: [m0, m, n, n0, n2, n1] → [m0, m, n, n + 1, n1, n0] flags: current, []

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%lep: ⊦ m ≤ m + n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: m0, m, n, n + 1, n1, n0