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Proof step 20 from proof div-ls-mone (previous step)

0 < j * (2 + i0) + j0, i0 < 0, 0 < j + j0, j0 < 0, i0 < j + j0, j0 < j, i0 < j, 0 < j, i0 < ~1 ⊦ i0 = ~1, j * (2 + i0) < 0 ∧ j0 < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
0 < j * (2 + i0) + j0, i0 < 0, 0 < j + j0, j0 < 0, i0 < j + j0, j0 < j, i0 < j, 0 < j, i0 < ~1, ~2 ≤ i0 ⊦ i0 = ~1

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j; (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%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% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ 0 < i → (i * j < 0 ↔ j < 0); (s)