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Proof step 4 from proof mod-neq-03 (previous step)

n0 + 1 < n ∧ n1 < n → (m0 * n + n0 + 1 = m * n + n1 ↔ m0 = m ∧ n0 + 1 = n1), m0 * n + n0 + 1 = m * n + n1, n1 < n, n0 < n, n1 < n0, n1 + 1 < n ⊦ n = 0, n0 = 0, n1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: modeq: ⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%rsucc-01: ⊦ (m + 1) * n = m * n + n;%ref-1: ⊦ ¬ n < n; from specification nat-basic1%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%l-06: ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; from specification nat-basic2%les: ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%lels-03: ⊦ m < n → m ≤ n; from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2