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Proof step 1 from proof zero-01 (conclusion)

⊦ i < 0 ∧ j < 0 → i + j * (n→i(n) + 1) < 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: nat2int-06: ⊦ ¬ n→i(n) + 1 < 0;%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%zero-gr-01: ⊦ i < 0 → (0 < i * j ↔ j < 0); from specification int-mult%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs