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Proof step 1 from proof zero-13 (conclusion)

⊦ i + j * n→i(n) < 0 ∧ m ≤ n ∧ 0 ≤ j → i + j * n→i(m) < 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: nat2int-04: ⊦ n→i(m) < n→i(n) ↔ m < n;%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%ls-pos: ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); from specification int-mult%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%ls-forward: ⊦ i * j < i * j0 → i ≠ 0; (forward) from specification int-mult%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs