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Proof step 6 from proof int2bits-lelen (previous step)

n→i(m) = i0 + 1, i < 2 ^ i1 * 2, 2 ^ i1 ≤ i, 0 ≤ i1, 0 < 2 ^ i1, 0 < i ⊦ # int2bits(i) ≤ m ↔ ¬ 2 ^ i0 ≤ i

User applied rule cut formula

And obtained the following new goals

Proof step 7:
n→i(m) = i0 + 1, i < 2 ^ i1 * 2, 2 ^ i1 ≤ i, 0 ≤ i1, 0 < 2 ^ i1, 0 < i ⊦ i = 0, # int2bits(i) ≤ m ↔ ¬ 2 ^ i0 ≤ i

Rule argument

fma: i = 0

Additional information

Used simplifier rules: int2bits-zero: ⊦ int2bits(0) = false ';% ⊦ # boolvar ' = 0 +1;% ⊦ 2 ^ i ≤ 0 ↔ i < 0;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ 0 ≤ i → ¬ 2 ^ i ≤ 0; Used formulas: i = 0