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Proof step 18 from proof div-minus-onediff-01 (previous step)

0 < j, i0 < ~1, i0 < 0 ⊦ 1 % j = 0, (i0 * j + 1) / j = i0 + 1, j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: div-add1-negpos-neg-01 seq: ⊦ i % j = 0 ∧ i + i0 < 0 ∧ i0 % j ≠ 0 ∧ i < 0 ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j + 1 subst: [j, i0, i] → [j, 1, i0 * j] flags: <all paths>

Additional information

Used simplifier rules: mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;%div-eq-zero-01: ⊦ 0 ≠ j → (i / j = 0 ↔ abs(i) < abs(j));%one-01: ⊦ i = 1 → j % i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s)% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s)%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ 0 < i → (i * j < 0 ↔ j < 0); (s)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i < 0 → (i < i * j ↔ j < 1); (s)% ⊦ i < 1 ↔ i ≤ 0; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < j → ¬ j < i; (s) Used terms: j, 1, i0 * j