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Proof step 6 from proof div-add-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mod-same: i ≠ 0 ⊦ i % i = 0;%div-same: i ≠ 0 ⊦ i / i = 1;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; 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-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs