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Proof step 5 from proof grz (previous step)

0 < i, 0 < i ^ j ⊦ i = 1, 0 ≤ j

User applied rule insert lemma

And obtained the following new goals

Proof step 6:
⊦ i ^ j = 1 ∧ 0 < j → abs(i) = 1
Proof step 7:
0 < i ^ j, 0 < i, i ^ ~ j = 1 ∧ 0 < ~ j → abs(i) = 1 ⊦ i = 1, 0 ≤ j

Rule argument

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

Additional information

Used terms: i, ~ j