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

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-ls-02: ⊦ 0 ≤ i → (i * j < 0 ↔ ¬ ¬(j < 0 ∧ i ≠ 0));%zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-ls-02: ⊦ 0 ≤ i → (i * j < 0 ↔ ¬ ¬(j < 0 ∧ i ≠ 0));%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs