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Proof step 154 from proof uint2bits-add-high (previous step)

1 = (1 - i1 * 2) + i1 * 2, (1 - i1 * 2) + 1 < i1, (1 - i1 * 2) + 1 < 4 * 2 ^ i2 + i1 * 2, 1 - i1 * 2 < i1, 1 - i1 * 2 < 4 * 2 ^ i2 + i1 * 2, 1 - i1 * 2 < 2 ^ i2, 1 - i1 * 2 < i2, 1 - i1 * 2 < 2 * (i1 + 1) + i2, 1 - i1 * 2 < 0, 1 - i1 * 2 < ~2, i1 + (1 - i1 * 2) < 0, 0 < i1, 0 ≤ i1, 0 ≤ 4 * 2 ^ i2 + i1 * 2, 0 < 4 * 2 ^ i2 + i1 * 2, 2 ^ i2 * 2 < i1, i1 * 2 < 4 * 2 ^ i2, 2 ^ i2 < i1, ~2 * 2 ^ i2 < i1 * 2, 2 < 2 * (i1 + 1) + i2, 2 * (2 ^ i2 + 1) < 2 * (i1 + 1) + i2, 2 < 4 * 2 ^ i2 + i1 * 2, 2 * (2 ^ i2 + 1) < 4 * 2 ^ i2 + i1 * 2, i1 * 2 ≤ 1 + 4 * 2 ^ i2, 2 * (2 ^ i2 + 1) < 4 * 2 ^ i2, 0 ≤ 2 ^ i2, 0 ≤ i2, 0 < 2 * (i1 + 1) + i2, 0 ≤ 2 * (i1 + 1) + i2, 1 < i1, 0 < 2 ^ i2, 1 - i1 * 2 < i1 + 4 * 2 ^ i2, i1 + (1 - i1 * 2) < 2 ^ i2, i1 + (1 - i1 * 2) < i2, i1 + (1 - i1 * 2) < 2 * (i1 + 1) + i2, (1 - i1 * 2) + 1 < i1 + 4 * 2 ^ i2, (i1 + (1 - i1 * 2)) + 1 < 2 ^ i2, 1 - i1 * 2 ≤ i1 + i2, ∀ i0, j. i→n(i0) < i→n(i2 + 2) → true → j < 0 ∧ 0 < i0 ∧ 2 ^ (i0 - 1) < abs(j) ∧ abs(j) ≤ 2 ^ i0 → uint2bits(j + 2 ^ i0) = sgn(int2bits(j).rest) ⊦ 4 * 2 ^ i2 = i1 * 2, uint2bits(2 ^ i2 * 2 - i1) + ((1 - i1 * 2) % 2 ≠ 0)' = sgn(int2bits(1 - i1 * 2).rest), ~1 = i2

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → ¬ j < i;%0 < i, i < j, j % i = 0 ⊦ (* i) * i0 < (* j) * j0 ↔ i0 < j / i * j0;% ⊦ 1 * i = i;% ⊦ i + j < (* i1) * i ↔ j < (i1 - 1) * i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ i + 0 = i;% ⊦ 0 < i ∧ 0 < j → 0 < i + j;% ⊦ i * j + i * j0 = i * (j + j0);% ⊦ i + i = 2 * i;% ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ (* i0) * i + i = (i0 + 1) * i;% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (j - i) = j;% ⊦ i * j = j * i;