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Proof step 10 from proof div-add-nosucc (previous step)

j = i1 + 1, j0 * j + j = i0 + 1, i0 / j = i1 / j + j0, 0 ≤ j0, 1 < j, 0 < j ⊦ i0 / j = j0, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: eq: ⊦ i = i + j ↔ j = 0; from specification int-basic2%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%zero-02: ⊦ 0 < i ∧ 0 < j → 0 < i + j; from specification int-basic2%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs