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Proof step 8 from proof int2bits-len-abs (previous step)

# int2bits(j) + 1 < # int2bits(~ j) ⊦ 2 ^ i + j ≠ 0 ∧ ¬ 0 ≤ j, j ≠ 2 ^ i0 ∧ ¬ j ≤ 0

User applied rule cut formula

And obtained the following new goals

Proof step 9:
j < 0, # int2bits(j) + 1 < # int2bits(~ j) ⊦ 2 ^ i + j ≠ 0 ∧ ¬ 0 ≤ j, j ≠ 2 ^ i0 ∧ ¬ j ≤ 0

Rule argument

fma: j < 0

Additional information

Used simplifier rules: int2bits-zero: ⊦ int2bits(0) = false ';%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 ↔ j ≤ i;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ # boolvar ' = 0 +1;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ 0 ≤ j → 2 ^ j ≠ 0;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # 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 < 0