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Proof step 2 from proof shiftzero-max-neg (previous step)

true, ∀ 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) ⊦ i < 0 ∧ # int2bits(i) ≤ n ∧ 0 < j ∧ n ≠ 0 → >>>(n, i, j) < 2 ^ (n→i(n) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# 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: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;