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Proof step 28 from proof bxor-flip-one-bit (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
# int2bits(i) = i→n(j), bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j))) = [], 0 ≤ j, 0 ≤ i, 2 ^ j * 2 < i ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ bits2int([]) = 0;