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Proof step 29 from proof shiftzero-max-neg-01 (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) → 0 < j0 ∧ i < 0 ∧ 2 ^ (n→i(n) - j0) ≤ >>>(n, i, j0) ∧ j0 ≤ n→i(n) ∧ # int2bits(i) ≤ n → n = 0 ⊦ n = 0, i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
# 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) → 0 < j0 ∧ i < 0 ∧ 2 ^ (n→i(n) - j0) ≤ >>>(n, i, j0) ∧ j0 ≤ n→i(n) ∧ # int2bits(i) ≤ n → n = 0 ⊦ i = ~1, n = 0

Rule argument

Additional information

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