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Proof step 1 from proof zero-02 (conclusion)

⊦ i * n→i(n) < 0 ↔ n ≠ 0 ∧ i < 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;%nat2int-08: ⊦ ¬ n→i(n) < 0;%zero-ls-02: ⊦ 0 ≤ i → (i * j < 0 ↔ ¬ ¬(j < 0 ∧ i ≠ 0)); from specification int-mult%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%zero-left: ⊦ 0 * i = 0; from specification int-mult%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%zero-ls-01: ⊦ i < 0 → (i * j < 0 ↔ 0 < j); from specification int-mult