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Proof step 101 from proof bxor-flip-one-bit (previous step)

# bits = i→n(j), int2bits(i) = bits0 + bits, 0 ≤ j, 0 ≤ i, 2 ^ j * 2 < i, bits ≠ [] → bitsxor(bits0 + bits, (false ' + true ') + mklist(false, i→n(j))) = bitsxor(bits0, false ' + true ') + bitsxor(bits, mklist(false, i→n(j))) ⊦ bits = [], bits0 = [], (int2bits(bits2int(bitsxor(bits0 + bits, false ' + true ' + mklist(false, i→n(j)))))[# bits0 - 1] ↔ ¬ bits0.last) ∧ int2bits(bits2int(bitsxor(bits0 + bits, false ' + true ' + mklist(false, i→n(j)))))[# bits0 - 1, bits0.last] = bits0 + bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 102:
# 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 = [], (int2bits(bits2int(bitsxor(bits0, false ' + true ') + bitsxor(bits, mklist(false, i→n(j)))))[# bits0 - 1] ↔ ¬ bits0.last) ∧ int2bits(bits2int(bitsxor(bits0, false ' + true ') + bitsxor(bits, mklist(false, i→n(j)))))[# bits0 - 1, bits0.last] = bits0 + bits

Rule argument

Additional information

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