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Proof step 7 from proof div-add1-posneg-pos-02 (previous step)

i0 + 1 = i1 + j, i % j = 0, i0 + 1 < j, 0 < j, 0 ≤ i + i0, 0 ≤ i, i0 < 0, i0 < j ⊦ (i + i0) / j = i / j + i1 / j, 1 < j ∧ i0 % j = 0, 1 < j ∧ i0 % j ≠ 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 3

Additional information

Used simplifier rules: one-01: ⊦ i = 1 → j % i = 0;%one: ⊦ i / 1 = i;%div-add1-negpos-pos: ⊦ i % j = 0 ∧ 0 ≤ i + i0 ∧ i < 0 ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j;%one-01: ⊦ i = 1 → j % i = 0;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2