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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;% ⊦ 0 ≤ i → (i * j < 0 ↔ j < 0 ∧ i ≠ 0); (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i * 0 = 0; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ 0 < n ↔ n ≠ 0; (s)% ⊦ i < 0 → (i * j < 0 ↔ 0 < j); (s)