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Proof step 48 from proof int2int-neg (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 49:
j = ~(2 ^ i), j + 2 ^ i = 0, j < 0, j < i + 2, j ≤ 0, j ≤ i + 2, j ≤ 4 * 2 ^ i, j < 4 * 2 ^ i, j < 3 * 2 ^ i, j ≤ 3 * 2 ^ i, j < 2 ^ i, j + 1 < 2 ^ i, j ≤ i, j + 1 < 4 * 2 ^ i, j < i, j ≤ 2 ^ i, 0 < 2 ^ i * 2 + j, 0 < 4 * 2 ^ i + j, 0 < 2 ^ i, ~2 ≤ i, 0 ≤ 4 * 2 ^ i + j, 0 ≤ i, 0 ≤ 2 ^ i, ~2 < i ⊦ 4 * 2 ^ i = 2 ^ (n→i(# uint2bits(4 * 2 ^ i + j)) - 1), 3 * 2 ^ i + j = 0

Rule argument

fma: j = ~(2 ^ i)

Additional information

Used simplifier rules: ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ~ i = j ↔ i + j = 0;% ⊦ i * j = j * i;% ⊦ (* i) ≤ j ∧ i ≠ 0 → i - 1 < j; (forward) Used formulas: j = ~(2 ^ i)