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Proof step 2 from proof abs-15 (previous step)

abs(j) < i, j < 0, i + j ≤ 0 ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%abs-08: ⊦ i < 0 → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%sub-ls: ⊦ i - j < i0 ↔ i < j + i0;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0;%add-c: ⊦ i + j = j + i; from specification int-basic2