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

i0 < abs(j), j < 0, 0 ≤ i0, 0 ≤ i ⊦ (i * abs(j) + i0) / j = ~ i

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i0 ∧ i0 < j ∧ 0 ≤ i → (i * j + i0) / j = i

Rule argument

spec: <toplevel> name: mult seq: ⊦ 0 ≤ i0 ∧ i0 < j ∧ 0 ≤ i → (i * j + i0) / j = i subst: [i0, j, i] → [i0, ~ j, i]

Additional information

Used simplifier rules: neg: j ≠ 0 ⊦ i / ~ j = ~(i / j);% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i * j = ~(i * j); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i + ~ j = i - j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ i < ~ j ↔ i + j < 0; (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i * 0 = 0; (s)% ⊦ i - 0 = i; (s)% ⊦ i + 0 = i; (s)% ⊦ i ≤ j → ¬ j < i; (s) Used terms: i0, ~ j, i