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Proof step 19 from proof shiftzero-max-neg (previous step)

0 ≤ 2 ^ n→i(n) - 1 ∧ 0 < j, 1 ≤ n, 0 < j, 2 ^ (n→i(n) - 1) ≤ (2 ^ n→i(n) - 1) / 2 >> (j - 1), ∀ n, i, j0. i→n(j0) < i→n(j) → true → i < 0 ∧ # int2bits(i) ≤ n ∧ 0 < j0 ∧ n ≠ 0 → >>>(n, i, j0) < 2 ^ (n→i(n) - 1) ⊦ j = 0, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
1 ≤ n, 0 < j, 2 ^ (n→i(n) - 1) ≤ (2 ^ n→i(n) - 1) / 2 >> (j - 1), ∀ n, i, j0. i→n(j0) < i→n(j) → true → i < 0 ∧ # int2bits(i) ≤ n ∧ 0 < j0 ∧ n ≠ 0 → >>>(n, i, j0) < 2 ^ (n→i(n) - 1) ⊦ j = 0, n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ n→i(n);% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;