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Proof step 36 from proof shiftzero-max-neg-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward)