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Proof step 18 from proof int2bits-lelen-07 (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: j = ~(2 ^ i)

Additional information

Used simplifier rules: int2bits-len-eq: ⊦ j ≠ 2 ^ n→i(# int2bits(j));%int2bits-2pot: 0 ≤ i ⊦ int2bits(2 ^ i) = false ' + true ' + mklist(false, i→n(i));%int2bits-2pot-01: 0 ≤ i ⊦ int2bits(~(2 ^ i)) = true ' + mklist(false, i→n(i));% ⊦ ~ i = j ↔ i + j = 0;% ⊦ ~ i + i = 0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ~ ~ i = i;% ⊦ i + ~ j = i - j;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + 0 = i;% ⊦ ~ i < 0 ↔ 0 < i;% ⊦ 0 < 2 ^ i ↔ 0 ≤ 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)