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Proof step 9 from proof int2bits-add-high (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
j < 0, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ i, j * 2 ^ i + j0 < 0 ⊦ sgn(int2bits(j * 2 ^ i + 2 ^ (n→i(# int2bits(~ j)) + i) + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0).rest)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 < i → n→i(i→n(i)) = i;% ⊦ n→i(m + n) = n→i(m) + n→i(n);