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Proof step 1 from proof abs-ls (conclusion)

⊦ i < j ∧ j < 0 → abs(j) < abs(i)

Heuristic simplifier applied rule simplifier

And closed the goal!

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;%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j;%sub-zero: ⊦ i + (0 - i0) = i - i0;%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j;%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%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1