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Proof step 10 from proof nls-mult-div (previous step)

i < (i * i1) / j ∧ (i * i1) / j < i → i < i, (i * i1) / j < i, 0 < j, i < (i * i1) / j, i1 < j, 1 < (i * i1) / j, 0 < (i * i1) / j, 0 < i ⊦ i = 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: irref: ⊦ ¬ i < i; from specification int-basic1