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Proof step 4 from proof square-one-le (previous step)

i * i ≤ 1 ⊦ i + 1 = 0, i = 1, i = 0

User applied rule apply lemma

And obtained the following new goals

Proof step 5:
⊦ i * i ≤ j * j ↔ abs(i) ≤ abs(j)
Proof step 6:
i * i ≤ 1 * 1 ↔ abs(i) ≤ abs(1), i * i ≤ 1 ⊦ i + 1 = 0, i = 0, i = 1

Rule argument

spec: <toplevel> name: square-le seq: ⊦ i * i ≤ j * j ↔ abs(i) ≤ abs(j) subst: [i, j] → [i, 1]

Additional information

Used terms: i, 1