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Proof step 1 from proof two-mult (conclusion)
⊦ 0 < i ∧ 0 < j0 → (2 * i - 1) / (2 * j0) = (i - 1) / j0
System applied rule normalize
And obtained the following new goals
Proof step 2
:
0 < i, 0 < j0 ⊦ (2 * i - 1) / (2 * j0) = (i - 1) / j0
Rule argument
Additional information