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Proof step 36 from proof shiftzero-max-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
# int2bits(i) ≤ n, i < ~1, i < 0, 2 ^ (n→i(n) - 1) ≤ (i - 1) / 2 + 2 ^ n→i(n - 1), ∀ n, i, j0. i→n(j0) < i→n(1) → true → i < 0 ∧ # int2bits(i) ≤ n ∧ 0 < j0 ∧ n ≠ 0 → >>>(n, i, j0) < 2 ^ (n→i(n) - 1) ⊦ i = ~1, n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ n -1 = n - 1;% ⊦ # bits ≤ n → # fillsgn(n, bits) = n;