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Proof step 20 from proof bxor-flip-one-bit (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ bits ≠ [] → (bits[0] ↔ bits.first) subst: [bits] → [int2bits(i)] rotate: F spec: list-del inst: bitlista name: get-zero

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ []; Used terms: int2bits(i)