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Proof step 5 from proof exin2pot-neg (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 6:
0 < 2 ^ i * 2 + j, 2 ^ i + j ≤ 0, 0 ≤ i, j < 0 ⊦ ~1 ≤ i ∧ j + 2 ^ i < 0 ∧ 0 ≤ 2 ^ (i + 1) + j, ∃ i. ~1 ≤ i ∧ j + 2 ^ i < 0 ∧ 0 ≤ 2 ^ (i + 1) + j

Rule argument

pos: right 1 terms: i abort: F discard: F computed: F

Additional information

Used terms: i