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Proof step 25 from proof bxor-flip-one-bit (previous step)

int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j))))) ≠ [], # int2bits(i) = i→n(j), 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j))))).first ∧ 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 26:
# 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)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ int2bits(j).first ↔ j < 0;% ⊦ int2bits(j) ≠ [];