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Proof step 7 from proof div-add-succ-neg (previous step)

1 < j, 0 < j, i < 0 ⊦ i % j = 0, i / j = ~((~ i + (j - 1)) / j) + 1, j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ 1 < j ∧ 0 < i ∧ i % j ≠ 0 → (i + (j - 1)) / j = i / j + 1

Rule argument

spec: <toplevel> name: div-add-succ seq: ⊦ 1 < j ∧ 0 < i ∧ i % j ≠ 0 → (i + (j - 1)) / j = i / j + 1 subst: [j, i] → [j, ~ i] flags: <all paths>

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%Div-left: ⊦ j ≠ 0 → ~ i / j = ~(i / j);% ⊦ 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) = j - i; (s)% ⊦ i + (j - i) = j; (s)% ⊦ i - j - i0 = i - j + i0; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ ~ i = 0 ↔ i = 0; (s) Used terms: j, ~ i