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Proof step 134 from proof bxor-flip-one-bit (previous step)

bits1 ≠ [] ∧ false ' = false ', int2bits(i) = (bits1 + boolvar ') + bits, # bits = i→n(j), bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = (bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))), 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ bits = [], bits1 = [], (sgn((bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))))[# bits1] ↔ ¬ boolvar) ∧ sgn((bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))))[# bits1, boolvar] = int2bits(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 135:
int2bits(i) = (bits1 + boolvar ') + bits, # bits = i→n(j), bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = (bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))), 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ bits = [], bits1 = [], (sgn((bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))))[# bits1] ↔ ¬ boolvar) ∧ sgn((bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))))[# bits1, boolvar] = int2bits(i)

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;