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Proof step 12 from proof int2bits-add-low (previous step)

j * 2 ^ i + j0 < ~1 ∧ 0 ≤ n→i(# int2bits(~(j * 2 ^ i + j0))), 0 < i, j < 0, j0 < 0, 0 < 2 ^ (i - 1) + j0, j * 2 ^ i + j0 < 0, ¬ # int2bits(j * 2 ^ i + j0) ≤ i→n(n→i(# int2bits(~(j * 2 ^ i + j0)))) + 1 ⊦ sgn(int2bits(2 ^ n→i(# int2bits(~(j * 2 ^ i + j0))) + j * 2 ^ i + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
1 + # int2bits(~(j * 2 ^ i + j0)) < # int2bits(j * 2 ^ i + j0), j * 2 ^ i + j0 < 0, 0 < 2 ^ (i - 1) + j0, j0 < 0, j < 0, 0 < i, j * 2 ^ i + j0 < ~1 ⊦ sgn(int2bits(2 ^ n→i(# int2bits(~(j * 2 ^ i + j0))) + j * 2 ^ i + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0))

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ i→n(n→i(n)) = n;% ⊦ 0 ≤ n→i(n);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;