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Proof step 2 from proof div-mult-num-01 (
previous step
)
j % j0 = 0, j ≠ 0, j0 ≠ 0 ⊦ (i * j0) / j = i / j / j0
Heuristic simplifier applied rule simplifier
And closed the goal!
Rule argument
Additional information
Used simplifier rules: div-div-switch: ⊦ j ≠ 0 ∧ i ≠ 0 ∧ i % j = 0 → i1 / i / j = (i1 * j) / i;