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Proof step 46 from proof bxor-flip-one-bit (previous step)

bits0 ≠ [], int2bits(i) = bits0 + bits, # bits = i→n(j), 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ 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

Rule argument

Additional information

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