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Proof step 35 from proof int2int-neg (previous step)

j ≤ 4 * 2 ^ i, j < 4 * 2 ^ i, j < 3 * 2 ^ i, j ≤ 3 * 2 ^ i, 2 ^ i + j ≤ 0, 0 < 2 ^ i * 2 + j, 2 ^ i + j ≤ i + 2, 2 ^ i + j < i + 2, j < i + 2, j < 0, j < 2 ^ i, j + 1 < 2 ^ i, j ≤ i, 0 ≤ i, j + 1 < 4 * 2 ^ i, 0 ≤ 4 * 2 ^ i + j, 0 < 4 * 2 ^ i + j, 2 ^ i + j ≤ i, j < i, j ≤ i + 2, j ≤ 0, j ≤ 2 ^ i, 0 ≤ 2 ^ i, 0 < 2 ^ i ⊦ j = ~(3 * 2 ^ i), 4 * 2 ^ i = 2 ^ (n→i(# uint2bits(4 * 2 ^ i + j)) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
j ≤ 4 * 2 ^ i, j < 4 * 2 ^ i, j < 3 * 2 ^ i, j ≤ 3 * 2 ^ i, 2 ^ i + j ≤ 0, 0 < 2 ^ i * 2 + j, 2 ^ i + j ≤ i + 2, 2 ^ i + j < i + 2, j < i + 2, j < 0, j < 2 ^ i, j + 1 < 2 ^ i, j ≤ i, 0 ≤ i, j + 1 < 4 * 2 ^ i, 0 ≤ 4 * 2 ^ i + j, 0 < 4 * 2 ^ i + j, 2 ^ i + j ≤ i, j < i, j ≤ i + 2, j ≤ 0, j ≤ 2 ^ i, 0 ≤ 2 ^ i, 0 < 2 ^ i ⊦ 4 * 2 ^ i = 2 ^ (n→i(# uint2bits(4 * 2 ^ i + j)) - 1), 3 * 2 ^ i + j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ~ i = j ↔ i + j = 0;