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Proof step 9 from proof two-le-three (previous step)

1 < i ^2, 0 < i ^2, 3 < i ^2, abs(i) ≤ 1, abs(i) < i ^2 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ i ≠ 0 ∧ i ≠ 1 → (abs(i) ≤ 1 ↔ i = ~1)
Proof step 11:
¬(i ≠ 0 ∧ i ≠ 1), 1 < i ^2, 0 < i ^2, 3 < i ^2, abs(i) ≤ 1, abs(i) < i ^2 ⊦
Proof step 19:
i ≠ 0 ∧ i ≠ 1, 1 < i ^2, 0 < i ^2, 3 < i ^2, i = ~1, abs(i) < i ^2 ⊦

Rule argument

spec: int-abs inst: name: abs-one-01 seq: ⊦ i ≠ 0 ∧ i ≠ 1 → (abs(i) ≤ 1 ↔ i = ~1) subst: [i] → [i] flags: rewrite

Additional information

Used terms: i