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Proof step 22 from proof bxor-flip-one-bit (previous step)

int2bits(i) ≠ [], # 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)))))[0] ↔ ¬ int2bits(i).first) ∧ int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j)))))[0, int2bits(i).first] = int2bits(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ ¬ i < j ↔ j ≤ i;%bits ≠ [] ⊦ int2bits(bits2int(bits)) = sgn(bits);% ⊦ int2bits(j).first ↔ j < 0;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ int2bits(j) ≠ [];