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Proof step 1 from proof ls-same-01 (conclusion)

⊦ 0 < i → (i < i * j ↔ 1 < j)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: mult-gt-one: ⊦ 0 < i ∧ 1 < i1 → i < i1 * i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j);%one-01: ⊦ i = 1 → i * j = j;%irref: ⊦ ¬ i < i; from specification int-basic1%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs