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Proof step 9 from proof shiftzero-max-neg-01 (previous step)

¬(i < ~1 ∧ # int2bits(i) ≤ n ∧ 1 < j), # int2bits(i) ≤ n, j ≤ n→i(n), 2 ^ (n→i(n) - j) ≤ >>>(n, i, j), i < 0, 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 10:
# int2bits(i) ≤ n, j ≤ n→i(n), 2 ^ (n→i(n) - j) ≤ >>>(n, i, j), i < 0, 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, i ≠ ~1 ∧ j ≠ 1

Rule argument

Additional information

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