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Proof step 27 from proof int2bits-add-high (previous step)

¬(j0 < 0 ∧ 0 < i ∧ 2 ^ (i - 1) < abs(j0) ∧ abs(j0) ≤ 2 ^ i ∧ 0 < 2 ^ n→i(# int2bits(~ j)) + j), 0 ≤ (2 ^ n→i(# int2bits(~ j)) + j) * 2 ^ i + j0, j < 0, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ i, j * 2 ^ i + j0 < 0 ⊦ sgn(uint2bits((2 ^ n→i(# int2bits(~ j)) + j) * 2 ^ i + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0).rest)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: int2bits-lelen-07: ⊦ ¬ 2 ^ n→i(# int2bits(~ j)) + j ≤ 0;% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ i < 0 ∧ j < 0 → i + j < 0;% ⊦ fillsgn(0, bits) = bits;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 1 * i = i;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ i < ~ j ↔ i + j < 0;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;