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Proof step 13 from proof shiftzero-max-neg (previous step)

~1 + 1 = 0, # true ' ≤ n, 2 ^ (n→i(n) - 1) ≤ >>>(n, ~1, j), 0 < j, ∀ 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) ⊦ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
2 ^ (n→i(n) - 1) ≤ >>>(n, ~1, j), 0 < j, 1 ≤ n, ∀ 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) ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ # boolvar ' = 0 +1;