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Proof step 1 from proof div-add-nosucc (conclusion)

⊦ 1 < j ∧ 0 ≤ i ∧ i % j = 0 → (i + (j - 1)) / j = i / j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i % j = 0, 0 < j, 1 < j, 0 ≤ i ⊦ (i + j - 1) / j = i / j

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) 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%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs