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Proof step 3 from proof square-le (previous step)

abs(j) < abs(i), i * i ≤ j * j ⊦

User applied rule cut formula

And closed the goal!

Rule argument

fma: j * j < i * i

Additional information

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