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Proof step 11 from proof square-ls (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: abs-eqall seq: ⊦ abs(i * j) = abs(i) * abs(j) subst: [j, i] → [i, i] flags: <rotated> <all paths>

Additional information

Used simplifier rules: square-abs: ⊦ abs(i * i) = i * i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: i, i