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Proof step 151 from proof bxor-flip-one-bit (previous step)

int2bits(i) = false ' + true ' + bits, # bits = i→n(j), bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = false ' + false ' + bitsxor(bits, mklist(false, i→n(j))), 0 < i, 0 ≤ j, 3 + i→n(j) ≤ # int2bits(i), false ∈ int2bits(i) ⊦ i = 0, # int2bits(i) = 2 + i→n(j), i = 2 ^ j * 2, bits = [], ¬ sgn(bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))))[1] ∧ sgn(bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))))[1, true] = int2bits(i)

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: discard

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m + n = m + n0 ↔ n = n0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;