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