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Proof step 10 from proof int2bits-len-abs (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: j = ~(2 ^ i)

Additional information

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