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Proof step 7 from proof div-add-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 < j, 1 < j, 0 < i, 0 ≤ j ⊦ i % j = 0, (i - 1) / j = i / j

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% ⊦ i + j = i + i0 ↔ j = i0; (s)% ⊦ 1 ≤ i ↔ 0 < i; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)