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Proof step 3 from proof square-ls-04 (previous step)

i * i + i < 2, 0 < i ⊦ 1 + i * i < 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i + j = j + i
Proof step 5:
i * i + i < 2, 0 < i ⊦ i * i + 1 < 2

Rule argument

spec: int-basic2 inst: name: add-c seq: ⊦ i + j = j + i subst: [j, i] → [i * i, 1] flags: rewrite

Additional information

Used terms: i * i, 1