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Proof step 16 from proof bxor-flip-one-bit (previous step)

0 ≤ i, 0 ≤ i0, 0 < i0, 0 < i, i < 2 ^ i0, 2 ^ (3 + i0) < i ⊦ ~2 = i0, ( int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(i0 + 2)))))[# int2bits(i) - i→n(i0 + 2) + 1] ↔ ¬ int2bits(i)[# int2bits(i) - i→n(i0 + 2) + 1]) ∧ int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(i0 + 2))))) [# int2bits(i) - i→n(i0 + 2) + 1, int2bits(i)[# int2bits(i) - i→n(i0 + 2) + 1] ] = int2bits(i)

User applied rule cut formula

And closed the goal!

Rule argument

fma: 2 ^ (3 + i0) < 2 ^ i0

Additional information

Used simplifier rules: ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;%0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0;% ⊦ i + j < j ↔ i < 0; Used formulas: 2 ^ (3 + i0) < 2 ^ i0