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Proof step 2 from proof le-one (previous step)

0 < j, 1 < j ⊦ i * j ≤ 1 ↔ i ≤ 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ls-same: ⊦ 0 < i → (i * j < i ↔ j < 1);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs