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Proof step 14 from proof exin2pot (previous step)

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, 0 ≤ i + 1 ∧ 2 ^ (i + 1) ≤ j ∧ j < 2 ^ ((i + 1) + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
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, 0 ≤ i + 1 ∧ 2 ^ (i + 1) ≤ j ∧ j < 2 ^ (2 + i)

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2