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Proof step 15 from proof int2bits-lslen-07 (previous step)

j + 2 ^ i = ~1, 3 + i→n(i) < # int2bits(j + 1), 0 ≤ 2 ^ i * 2 + j, 0 ≤ i, j < 0 ⊦

User applied rule cut formula

And closed the goal!

Rule argument

fma: j = ~(2 ^ i) - 1

Additional information

Used simplifier rules: int2bits-2pot-01: 0 ≤ i ⊦ int2bits(~(2 ^ i)) = true ' + mklist(false, i→n(i));% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ ~ i = j ↔ i + j = 0;% ⊦ ~ i - j = ~(i + j);% ⊦ i + ~ j = i - j;% ⊦ i - j + i = ~ j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ # mklist(boolvar, n) = n;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used formulas: j = ~(2 ^ i) - 1