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Proof step 71 from proof int2bits-add-low (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 72:
0 ≤ j0 + 2 ^ (i - 1), 0 < 2 ^ (i - 1) + j0, j0 < 0, j < 0, 0 < i, j * 2 ^ i + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ i, 0 < 2 ^ n→i(# int2bits(1 - j)) + j, 0 ≤ (2 ^ n→i(# int2bits(1 - j)) + j) * 2 ^ i + j0 ⊦ j * 2 ^ i + j0 = 0, sgn(int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0))) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0)), ∃ i0. 1 = 2 ^ i0 + j

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i + 0 = i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i + ~ j = i - j;% ⊦ ~(i - 1) = ~ i + 1;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;