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Proof step 6 from proof square-le-02 (previous step)

i * i < abs(i), abs(i) * abs(i) < abs(i) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ 0 < i → (i * j < i ↔ j < 1)

Rule argument

spec: <toplevel> name: ls-same seq: ⊦ 0 < i → (i * j < i ↔ j < 1) subst: [j, i] → [abs(i), abs(i)] flags: <all paths>

Additional information

Used simplifier rules: associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%square-lez: ⊦ ¬ i * i < 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1% ⊦ i +1 = i + 1; (s)% ⊦ i + j < j ↔ i < 0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: abs(i), abs(i)