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Proof step 3 from proof abs-eqall (previous step)

i < 0 ⊦ i * abs(j) + abs(i * j) = 0

User applied rule cut formula

And closed the goal!

Rule argument

fma: j < 0

Additional information

Used simplifier rules: minus-left: ⊦ ~ i * j = ~(i * j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-ge-01: ⊦ i < 0 → (0 ≤ i * j ↔ j ≤ 0);%zero-le-01: ⊦ i < 0 → (i * j ≤ 0 ↔ 0 ≤ j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs%abseq: ⊦ abs(i) + i = 0 ↔ i ≤ 0; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%abs-18: ⊦ abs(i) = i ↔ 0 ≤ i; from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used formulas: j < 0