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Proof step 2 from proof numabs-01 (previous step)

0 ≤ j, 0 ≤ j0 ⊦ 0 < i → abs(i * j + j0) = i * j + j0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%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%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%abs-18: ⊦ abs(i) = i ↔ 0 ≤ i; from specification int-abs