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Proof step 55 from proof bxor-flip-one-bit (previous step)

2 < i, 0 < i ⊦ i = 0, (sgn(bitsxor(int2bits(i).butlast, false ') + (¬ int2bits(i).last)')[# int2bits(i) - 1] ↔ ¬ int2bits(i).last) ∧ sgn(bitsxor(int2bits(i).butlast, false ') + (¬ int2bits(i).last)')[# int2bits(i) - 1, int2bits(i).last] = int2bits(i)

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [int2bits(i), boolvar, bits] rotate: T spec: list-last inst: bitlista name: last-elim

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];%bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar '; (elim) Used terms: int2bits(i), boolvar, bits