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Proof step 2 from proof exin2pot-neg (previous step)

j < 0 ⊦ ∃ i. ~1 ≤ i ∧ j + 2 ^ i < 0 ∧ 0 ≤ 2 ^ (i + 1) + j

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < j → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1))
Proof step 4:
0 < ~ j → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ ~ j ∧ ~ j < 2 ^ (i + 1)), j < 0 ⊦ ∃ i. ~1 ≤ i ∧ j + 2 ^ i < 0 ∧ 0 ≤ 2 ^ (i + 1) + j

Rule argument

spec: int-pot inst: name: exin2pot seq: ⊦ 0 < j → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1)) subst: [j] → [~ j] flags: current, []

Additional information

Used terms: ~ j