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Proof step 9 from proof uint2bits-2pot-04 (previous step)

2 ^ (n→i(# bits) + 1) = i + 1, 0 ≤ i ⊦ uint2bits(i) = false ' + true ' + bits, false ∈ bits

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = 2 ^ (n→i(# bits) + 1) - 1

Additional information

Used simplifier rules: uint2bits-2pot-02: 0 < i ⊦ uint2bits(2 ^ i - 1) = false ' + mklist(true, i→n(i));% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + (j - i) = j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ n→i(n);% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ i→n(n→i(n) + 1) = n + 1;% ⊦ mklist(boolvar, m + n) = mklist(boolvar, m) + mklist(boolvar, n);% ⊦ ¬ false ∈ bits → mklist(true, # bits) = bits;% ⊦ mklist(boolvar, 1) = boolvar ';% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ ¬ false ∈ bits → (bits + true ' = true ' + bits ↔ true);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used formulas: i = 2 ^ (n→i(# bits) + 1) - 1