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Proof step 12 from proof exin2pot (previous step)

∃ i. 0 ≤ i ∧ 2 ^ i ≤ j / 2 ∧ j / 2 < 2 ^ (i + 1), 0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ j0 ∧ j0 < 2 ^ (i + 1)) ⊦ j = 1, j = 0, ∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
0 ≤ i, 2 ^ i ≤ j / 2, j / 2 < 2 ^ (i + 1), 0 < j, 2 ^ i < 2 ^ (i + 1), ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ j0 ∧ j0 < 2 ^ (i + 1)) ⊦ j = 1, j = 0, ∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1)

Rule argument

Additional information

Used simplifier rules: forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs