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Proof step 9 from proof exin2pot (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 10:
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, 0 < j / 2, ∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1)
Proof step 13:
0 < j, ∃ i. 0 ≤ i ∧ 2 ^ i ≤ j / 2 ∧ j / 2 < 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

subst: [j0] → [j / 2]

Additional information

Used simplifier rules: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); (s)%j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j); (s)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < i → ¬ i ≤ i / 2; (s) Used terms: j / 2