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Proof step 4 from proof bxor-flip-one-bit (previous step)

2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ (int2bits(bits2int(bitsxor(int2bits(i), int2bits(2 ^ j))))[# int2bits(i) - i→n(j) + 1] ↔ ¬ int2bits(i)[# int2bits(i) - i→n(j) + 1]) ∧ int2bits(bits2int(bitsxor(int2bits(i), int2bits(2 ^ j))))[# int2bits(i) - i→n(j) + 1, int2bits(i)[# int2bits(i) - i→n(j) + 1]] = int2bits(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);%0 ≤ i ⊦ int2bits(2 ^ i) = false ' + true ' + mklist(false, i→n(i));% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;