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Proof step 2 from proof bxor-flip-one-bit (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ bxor(i, j) = bits2int(bitsxor(int2bits(i), int2bits(j)))
Proof step 4:
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)

Rule argument

name: bxor-def rotate: F subst: [j, i] → [2 ^ j, i]

Additional information

Used terms: 2 ^ j, i