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Proof step 3 from proof absls-02 (previous step)

i < 0 ⊦ abs(i) < abs(j) ↔ j < i, 0 < i, 0 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%sub-ls: ⊦ i - j < i0 ↔ i < j + i0;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0;%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1