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Proof step 2 from proof numabs (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: numabs-01 seq: 0 ≤ j, 0 ≤ j0 ⊦ 0 < i → abs(i * (* j) + (* j0)) = i * j + j0 subst: [j0, j, i] → [j0, j, i] flags: <all paths>

Additional information

Used simplifier rules: Mult-zero: ⊦ i * 0 = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ abs(i) = i ↔ 0 ≤ i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i + 0 = i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ ¬ i < i; (s) Used terms: j0, j, i