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Proof step 6 from proof le-same-03 (previous step)

i * 1 ≤ i * j ↔ j ≤ 1, i < 0 ⊦ i ≤ i * j ↔ j ≤ 1

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;