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Proof step 6 from proof square-ls (previous step)

i * i < j * j ⊦ abs(i) < abs(j), 0 < abs(j) ∧ 0 < abs(j) ∧ abs(j) ≤ abs(i) ∧ abs(j) ≤ abs(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i * i < j * j ⊦ j = 0, abs(i) < abs(j), abs(j) ≤ abs(i)

Rule argument

Additional information

Used simplifier rules: associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%square-lez: ⊦ ¬ i * i < 0;% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)