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Proof step 2 from proof square-ls (
previous step
)
⊦ i * i < j * j ↔ abs(i) < abs(j)
User applied rule case distinction
And obtained the following new goals
Proof step 3
:
i * i < j * j ⊦ abs(i) < abs(j)
Proof step 10
:
abs(i) < abs(j) ⊦ i * i < j * j
Rule argument
pos: right 1
Additional information