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

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;%sub-add-01: ⊦ i + (j - i0) = i + j - i0;%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j;%lf-04: ⊦ ¬ j < i ∧ j < i0 → i < i0; (forward) from specification int-basic1%lf-07: ⊦ ¬ i < i0 ∧ j < i0 → j < i; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ 0 + i = i; (s)% ⊦ i +1 = i + 1; (s)%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