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Proof step 113 from proof bxor-flip-one-bit (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 114:
# bits = i→n(j), int2bits(i) = bits0 + bits, bitsxor(bits0 + bits, (false ' + true ') + mklist(false, i→n(j))) = bitsxor(bits0, false ' + true ') + bitsxor(bits, mklist(false, i→n(j))), bits0 = bits0.first ', 0 ≤ j, 0 ≤ i, 2 ^ j * 2 < i ⊦ bits = [], bits0 = [], bitsxor(bits0, false ' + true ') + bitsxor(bits, mklist(false, i→n(j))) = [], (sgn(bitsxor(bits0, false ' + true ') + bitsxor(bits, mklist(false, i→n(j))))[0] ↔ ¬ bits0.first) ∧ sgn(bitsxor(bits0, false ' + true ') + bitsxor(bits, mklist(false, i→n(j))))[0, bits0.first] = bits0 + bits

Rule argument

Additional information

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