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Proof step 10 from proof int-simp-04 (previous step)

0 < i0, 0 < i1 +1, i0 + i1 +1 < 0, ∀ i0. ¬(i0 + i1 < 0 ∧ 0 < i1 ∧ 0 < i0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i0 + i1 < ~1, 0 < i0, i0 + i1 < 0, ∀ i. ¬ i + i1 < 0 ∨ ¬ 0 < i1 ∨ ¬ 0 < i ⊦ i1 < 0

Rule argument

Additional information

Used simplifier rules: numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j;%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;%mone-06: ⊦ ~1 < i ↔ 0 ≤ i;%le-def: ⊦ i ≤ j ↔ ¬ j < i;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j;%numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i +1 = i + 1; (s)