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Proof step 4 from proof absls-02 (previous step)

i < 0 ⊦ 0 < i + abs(j) ↔ j < i, 0 < j

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: j < 0

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;%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j;%lst: ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i; from specification int-basic1%lst-02: ⊦ ¬ i < j → (¬ j < i ↔ i = j); from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used formulas: j < 0