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Proof step 6 from proof two-le-three (previous step)

i * i < 2 * 2, i * i ≤ 3, 1 < abs(i) ⊦ i = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ i * i < j * j ↔ abs(i) < abs(j)

Rule argument

spec: int-mult inst: name: square-ls seq: ⊦ i * i < j * j ↔ abs(i) < abs(j) subst: [j, i] → [2, i] flags: rewrite

Additional information

Used simplifier rules: num-between: i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j; from specification int-basic2 Used terms: 2, i