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Proof step 6 from proof pnle-pos (previous step)

i0 * j0 ≤ i * j, j < j0, i ≤ i0, 0 < j0, 0 ≤ j, 0 ≤ i, 0 < i0 ⊦ i0 = 0, 0 < i ∧ 0 < j

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-le: ⊦ 0 < i → (i * j ≤ 0 ↔ j ≤ 0);%zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-le: ⊦ 0 < i → (i * j ≤ 0 ↔ j ≤ 0);%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs