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Proof step 7 from proof div2-nls (previous step)

i + j = i0 * 2 + j0 ∧ j0 ≤ 0 ∧ abs(j0) < 2 ∧ i0 ≤ 0 ∧ 0 < 2, j < i0, i ≤ j, i < i0, i + j < 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i + j = i0 * 2 + j0, i < 0, i < i0, j < 0, i0 ≤ 0, j0 ≤ 0, i ≤ j, j < i0, abs(j0) < 2 ⊦

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%zero-01: ⊦ i < 0 ∧ j < 0 → i + j < 0; from specification int-basic2%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs