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Proof step 4 from proof absa (previous step)

i + j < 0 ⊦ 0 = (abs(i) + abs(j)) + i + j, 0 < i, 0 < j

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

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