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Proof step 14 from proof shiftzero-max-neg-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
j ≤ n→i(n), 2 ^ (n→i(n) - j) ≤ >>>(n, ~1, j), 0 < j, 1 ≤ n, ∀ 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 = 0

Rule argument

Additional information

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