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Proof step 1 from proof nummult-ls-03-01 (conclusion)
0 < i, j < 0 ⊦ (* j) < (* i) * i0 ↔ (j + 1 - i) / i < i0
Special step normalize
And obtained the following new goals
Proof step 2
:
0 < i, j < 0 ⊦ j < i * i0 ↔ (j + 1 - i) / i < i0