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Proof step 7 from proof square-le-06 (previous step)

i * i ≤ i + 1 ⊦ i < 0, i = 0, i = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i * i ≤ i + 1, 0 < i ⊦ i = 1

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1