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Proof step 3 from proof abs-add-le-zero (previous step)

i < 0 ⊦ 0 < abs(i) + abs(j), i = 0 ∧ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i < 0 ⊦ i = 0, i < abs(j)

Rule argument

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%sub-zero: ⊦ i + (0 - i0) = i - i0;%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0;%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2