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Proof step 1 from proof numls (conclusion)
⊦ (* i1) * j < (* i2) * j ↔ 0 < (i2 - i1) * j
Special step normalize
And obtained the following new goals
Proof step 2
:
⊦ i1 * j < i2 * j ↔ 0 < (i2 - i1) * j