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Proof step 1 from proof num-div-denom-01 (conclusion)
⊦ i1 ≠ 0 ∧ j ≠ 0 → (rational(i, i1).num * j) / rational(i, i1).denom = (i * j) / i1
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
⊦ i1 = 0, j = 0, (rational(i, i1).num * j) / rational(i, i1).denom = (i * j) / i1
Rule argument
Additional information