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Proof step 2 from proof eqone-02 (previous step)

i ^ ~ j = 1, j < 0 ⊦ abs(i) = 1

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ i ^ j = 1 ∧ 0 < j → abs(i) = 1

Rule argument

spec: <toplevel> name: eqone-01 seq: ⊦ i ^ j = 1 ∧ 0 < j → abs(i) = 1 subst: [i, j] → [i, ~ j]

Additional information

Used simplifier rules: ⊦ 0 < ~ i ↔ i < 0; (s) Used terms: i, ~ j