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Proof step 111 from proof cast-mod-neg-high (previous step)

int2bits(j % 2 ^ i) ≠ [], i→n(i) = (# int2bits(j % 2 ^ i))-1, j % 2 ^ i < ~1, i→n(i) < # int2bits(j % 2 ^ i), j + 2 ^ i < i, j + 2 ^ i ≤ 0, j < j % 2 ^ i, 0 ≤ 2 ^ (i - 1) + j % 2 ^ i, j < 0, 0 < i, j < 2 ^ (i - 1), 2 ^ (i - 1) + j < 0, j % 2 ^ i < 0 ⊦ j = ~1, i = 1, j % 2 ^ i = 0, 2 ^ (n→i(# int2bits(j % 2 ^ i)) - 1) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 112:
i→n(i) = # int2bits(j % 2 ^ i) - 1, j % 2 ^ i < 0, 2 ^ (i - 1) + j < 0, j < 2 ^ (i - 1), 0 < i, j < 0, 0 ≤ 2 ^ (i - 1) + j % 2 ^ i, j < j % 2 ^ i, j + 2 ^ i ≤ 0, j + 2 ^ i < i, i→n(i) < # int2bits(j % 2 ^ i), j % 2 ^ i < ~1 ⊦ 2 ^ (n→i(# int2bits(j % 2 ^ i)) - 1) = 0, j % 2 ^ i = 0, i = 1, j = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m ≤ m;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n + 1 ↔ ¬ n < m;%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ int2bits(j) ≠ [];% ⊦ n -1 = n - 1;