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Proof step 67 from proof bxor-flip-one-bit (previous step)

int2bits(i) = bits + boolvar ', bits = bits.first ', 2 < i, 0 < i, boolvar ∈ int2bits(i) ⊦ i = 0, bits = [], (sgn(bits + (¬ boolvar)')[# int2bits(i) - 1] ↔ ¬ boolvar) ∧ sgn(bits + (¬ boolvar)')[# int2bits(i) - 1, boolvar] = int2bits(i)

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

seq: ⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0) subst: [bits, boolvar, bits0] → [bits, boolvar0, bits0] rotate: T spec: list-data inst: bitlista name: elim-1

Additional information

Used simplifier rules: ⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0); (elim) Used terms: bits, boolvar0, bits0