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Proof step 5 from proof abs-eqall (previous step)

0 ≤ i ⊦ abs(i * j) = i * abs(j)

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-le-01: ⊦ i < 0 → (i * j ≤ 0 ↔ 0 ≤ j);%zero-ge-02: ⊦ 0 ≤ i → (0 ≤ i * j ↔ ¬ ¬(i = 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%abs-18: ⊦ abs(i) = i ↔ 0 ≤ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%abseq: ⊦ abs(i) + i = 0 ↔ i ≤ 0; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; 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 Used formulas: j < 0