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Proof step 2 from proof square-le-01 (previous step)

i * i < i ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < i → (i * j < i ↔ j < 1)

Rule argument

spec: int-mult inst: name: ls-same seq: ⊦ 0 < i → (i * j < i ↔ j < 1) subst: [j, i] → [i, i] flags: rewrite, current

Additional information

Used simplifier rules: zero-ls-03: ⊦ i ≤ 0 → (i * j < 0 ↔ ¬ ¬(0 < j ∧ i ≠ 0));%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%num-between: i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs Used terms: i, i