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Proof step 150 from proof bxor-flip-one-bit (previous step)

¬ bits0 ≠ [], bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = (false ' + bits0) + (¬ boolvar)' + bitsxor(bits, mklist(false, i→n(j))), # bits = i→n(j), int2bits(i) = (false ' + bits0) + boolvar ' + bits, false ∈ bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))), false ∈ int2bits(i), 3 + i→n(j) ≤ # int2bits(i), 0 ≤ j, 0 < i ⊦ bits = [], i = 2 ^ j * 2, # int2bits(i) = 2 + i→n(j), i = 0, (bits0 + (¬ boolvar)').first, (sgn(bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))))[1 + # bits0] ↔ ¬ boolvar) ∧ sgn(bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))))[1 + # bits0, boolvar] = int2bits(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 151:
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)

Rule argument

Additional information

Used simplifier rules: ⊦ bits = bits0 + boolvar ' + bits1 → boolvar ∈ bits;% ⊦ # [] = 0;% ⊦ boolvar '.first ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;