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Proof step 1 from proof longin-mult-mod (conclusion)
⊦ i ∈bint ∧ j ∈bint ∧ j0 ≠ 0 → ((i * j) % j0)∈long
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
j ∈bint, i ∈bint ⊦ j0 = 0, ((i * j) % j0)∈long
Rule argument
Additional information