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Proof step 5 from proof div-elim1 (previous step)

j < 0, 0 ≤ i, 0 < ~ j ∧ 0 ≤ i → (~ i0 = i / ~ j ∧ j0 = i % ~ j ↔ i = ~ i0 * ~ j + j0 ∧ 0 ≤ j0 ∧ j0 < ~ j) ⊦ i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 + j < 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

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