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Proof step 1 from proof nummult-ls-02 (conclusion)
⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0)
Special step normalize
And obtained the following new goals
Proof step 2
:
⊦ 0 < i ∧ 0 ≤ j → (j < i * i0 ↔ j / i < i0)