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Proof step 5 from proof shiftzero-max-neg (previous step)

¬(i < ~1 ∧ n < # int2bits(i) ∧ 1 < j), # int2bits(i) ≤ n, 2 ^ (n→i(n) - 1) ≤ >>>(n, i, j), i < 0, 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 6:
# int2bits(i) ≤ n, 2 ^ (n→i(n) - 1) ≤ >>>(n, i, j), i < 0, 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

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false);% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ ¬ i < j ↔ j ≤ i;